operation management full without cutting mark

OPERATION MANAGEMENT

ANDINFORMATION

SYSTEMS

INTERMEDIATE

GROUP - IPAPER 9

The Institute of Cost and Works Accountants of India12, SUNDER STREET, KOLKATA - 700 016

First Edition : January 2008

Published by:Director of StudiesThe Institute of Cost and Works Accountants of India12, SUDDER STREET, KOLKATA - 700 016

Printed at : Repro India Limited,50/2, TTC MIDC Industrial Area, Mahape, Navi Mumbai - 400 710, India

Copyright of these Study Notes in reserved by the Institute of Cost and Works Accountantsof India and prior permission from the Institute is necessary for reproduction of the whole orany part thereof.

SYLLABUSPAPER 9: OPERATION MANAGEMENT AND INFORMATION SYSTEMS

(ONE PAPER: 3 HOURS: 100 MARKS)

Section A: Operation Management (50 marks)1. Overview of Production Process

• Fabrication process• Metal working process – Forming, Heat Treatment, Welding, Surface treatment etc.• Machining process• Class of machine – Lathes, Drilling, Grinding, Milling, Plaining, Shaping, Slotting etc.• Special purpose machine – Special Grinding, Hobbing, Honing, Cutting Tools, Jigs and Fixtures etc.• Pump, Motor, Transformer, Electrical drives,• Classification of industries based on production process• Technological aspects of different production process like power, pollution control, recovery process,• Plant layout , Material handling system etc

2. Production Planning & Productivity Management• Time Study, Work Study, Method Study, Job Evaluation.• Production Planning and Control-Introduction.• Forecasting• Capacity Planning and Utilization.• Process Planning,• Project Planning.• Progressing and Follow-Up.• Dispatching.• Scheduling Technique & Line Balancing Problem• Economic Batch Production• Human Resource Planning• Material Requirement Planning• Productivity Measurement Techniques of Factors of Production• Quality Control

3. Maintenance Management• Obsolesce, Replacement of Machinery• Breakdown Maintenance, Preventive Maintenance & Routine Maintenance• Maintenance Techniques• Maintenance Organization• Maintenance Problems etc

4. Resource Management• Input-Output Ratio• Linear Programming• Transportation• Replacement of Machine• Change of Technology and its implication

Operation Management

CONTENTS

Study Note 1 Page No.

1. Overview of Production Process 1

1.0 Fabrication Process 2

Metal working process, Forming, Heat treatment,Welding, Surface treatment

1.1 Machining Process 5

Blanking, Boring, Broaching, Die Cutting, Drilling,Flame cutting or gas cutting, Grinding, etc.

1.2 Class of machine 7

Lathes, drilling, grinding, milling, plaining, shaping, slotting etc.

1.3 Special purpose machine 8

special grinding, hobbing, honing, cutting tools, jigs and fixtures etc.

1.4 Pump, motor, transformer, electrical drives 20

1.5 Classification of industries based on production process 23

1.6 Technological aspects of different production process like power,pollution control, recovery process 25

1.7 Plant layout 28

Study Note 2

2. Production Planning and Productivity Management 53

2.0 Time study, Work study, Method study, Job Evaluation. 54

2.1 Production planning and Control-Introduction 61

2.2 Forecasting 69

2.3 Capacity planning and Utilization 86

2.4 Process planning 120

2.5 Project planning 133

Page No.

2.6 Progressing and Follow-up 139

2.7 Dispatching 139

2.8 Scheduling Technique & Line Balancing Problem 140

2.9 Economic batch Production 157

2.10 Human Resource Planning 165

2.11 Material Requirement Planning 167

2.12 Productivity Measurement Techniques of Factors of Production 168

2.13 Quality Control 197

Study Note 3

3. Maintenance Management 219

3.0 Obsolescence, replacement of Machinery 220

3.1 Breakdown maintenance, preventive maintenanceand Routine Maintenance - Breakdown Maintenance 220

3.2 Maintenance Techniques 222

3.3 Maintenance Organization 222

3.4 Maintenance Problems 223

Study Note 4

4. Resource Management 229

4.0 Input-output ratio 230

4.1 Linear Programming 231

4.2 Transportation 251

4.3 Replacement of Machine and other Relevant Concepts 268

4.4 Change of Technology and its Implication 279

5. Objective type Questions and Answers 289

Information System

CONTENTS

Page No.

Study Note 1

1. Information System 317

1.1 Contribution of Information Technology in the Social lifeand Industrial fields 318

1.2 Information 319

1.3 Information System 319

1.4 Open and closed System 321

1.5 Evolution of Information system 321

1.6 Growth of Business Information System 324

1.7 Strategic Planning for Development of an Information System 325

1.8 Information system Infrastructure 327

1.9 Information system Organization 329

1.10 Information Systems Personnel Management 331

1.11 Quality of a Business Information System 331

1.12 Business Opportunities due to Development of Computers 332

Study Note 2

2. Hardware 3352.1 Hardware 336

2.2 Analog, Digital and Hybrid Computer 336

2.3 Evolution of Hardware 337

2.4 Different Generations of Computers? 338

2.5 How a Computer Functions 339

2.6 System of Stroge in Memory 342

2.7 Different Types of Memory 343

2.8 Secondary Storage Device 344

2.9 Virtual Memory 344

2.10 Classification of Digital Computers 344

2.11 Input / Output Devices Input Devices 346

2.12 Different Types of Input Devices 346

Page No.

2.13 Different types of Output Devices 348

2.14 Selection of a Computer System 353

2.15 Hardware Procurement 354

2.16 Installation of a Computer System 355

2.17 Hardware Manitenance 355

Study Note 3

3. Software and it’s Tools 3573.1 Software 358

3.2 Computer Program 358

3.3 Compiler 361

3.4 Interpreter 362

3.5 Difference between Compiler and Interpreter 362

3.7 Fourth Generation Language (4 GL) 365

3.8 Some General Software Features 365

3.9 Advanced Programming Languages 365

3.10 Visual Programming Language 366

3.11 Object Oriented Programming 366

3.12 Oracle 367

3.14 Application Software 368

3.15 Applicability of Program Language 369

Study Note 4

4. System Development Cycle 3714.1 System Development 372

4.2 Development of Prototype 372

4.3 System Development Life Cycle 372

4.4 System Development Team 373

4.5 System Proposal 373

4.6 System Study 376

4.7 System Design 376

4.8 Program Development 382

4.9 System Implementation 388

4.10 System Evaluation & Review 390

4.11 System Maintenance 390

Page No.

4.12 System Documentation 391

4.13 System Development Controls 392

4.14 Software Development Cost 392

4.15 Operating Cost 392

4.16 Readymade Software 393

4.17 Tailor made Software 394

Study Note 5

5. Data Management & Data Processing 395

5.1 File 396

5.2 Data Structure 396

5.3 Types of Files 396

5.4 File Organization 397

5.5 Data base 399

5.6 Database Management System (DBMS) 400

5.7 DBMS System Structure 402

5.8 Normalisation 405

5.9 Code’s Rules 405

5.10 Data Processing 406

5.11 Data Processing Activities 406

5.12 Factors for Decision on Mode of Data Processing 407

5.13 Advantage of Data Processing 407

5.14 Type of Data Processing 408

5.15 Information System Architecture 410

Study Note 6

6. Communication 413

6.1 Communication 414

6.2 Protocol 414

6.3 Modes of data transmission over a channel 415

Study Note 7

7. Information Management & Decision Making 4397.1 Data Source 440

7.2 Data Quality 440

7.3 Data Storage and Management 440

Page No.

7.4 Data Warehousing 441

7.5 Data Mart 442

7.6 Data Mining 442

7.7 Knowledge Management 443

7.8 Business Process Re-engineering (BPR) 445

7.9 Expert System 447

7.10 Artificial Intelligence (AI) 447

7.11 Decision Making Process 447

7.12 Decision Support system (DSS) 452

7.13 Flow of Information for supporting decision making 453

Study Note 8

8. Management Information System 455

8.1 Management Information System 456

8.2 Objectives of MIS 456

8.3 Basic Fetures of an MIS 456

8.4 Implementation of MIS 457

8.5 Different Approaches for developing Management Information System 458

8.6 Constraints in operating MIS 459

8.7 Limitations 459

8.8 Information Requirements of Management 460

8.9 Information need for Strategic Decision Making 462

8.10 Different types of Reports 463

8.11 Management Decision making 464

8.12 Components of Business Information System 465

8.13 Need for Integration of Information Enterprise - wide 466

8.14 Models used for representing the Information 468

8.15 Analysis of Information 468

8.16 Information requirement for Product Pricing Strategy 469

8.17 Executive Information System 469

Study Note 9

9. Enterprise Resource Management 4719.1 Introduction 472

9.2 Selection process of an ERP Package 473

Page No.

Study Note 10

10. Information System Control 481

10.2 Information System and Management Control 483

10.3 Controls in Information System 485

10.4 Controls in Processing Environment 487

10.5 Documentation Control 489

10.6 Computer Fraud 489

10.7 Security of Information System 490

10.8 Disaster Recovery 493

Study Note 11

11. Information System Audit 495

11.1 Objective 496

11.2 Definition 496

11.3 Internal Control 496

11.4 Audit Planning 498

11.5 Documentation 498

11.6 Testing 499

11.7 Outsourcing 502

11.8 Internal Auditors vs. External Auditors 503

11.9 Audit Charter 503

11.10 Information System Audit and Control Association (ISACA) 503

11.11 Pratical Examples 504

11.12 Short Notes 510

12. Revisional Questions 515

13. Suggested Answer to Revisional Questions 525

Part - IOperation

Management

Overview ofProduction

Process

STUDY NOTE - 1

Overview of Production Process

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1.0 Fabrication Process

Fabrication is the process of forming, casting, machining and welding of metals.

A fabricating process modifies the physical characteristics of materials with the help of certain labour ormachine operations. It is generally applied in metal working industries. The fabricating process involveschanging the shape of materials and connecting the parts. Fabrication work involves several operationslike forging, gas welding, flame-cutting, etc.

Classification of Fabrication processes:

(A) Fabrication process concerning Iron Smelting: Metals are widely used for fabrication purposes inmost of the modern industries. Various metals like iron have-to be smelted to get them in virgin form.Iron is obtained in free state as oxide and it is smelted in a Blast Furnace with the help of coke andlimestone.

(B) Fabrication Process concerning Making of Steel: Steel making involves re-melting the pig iron andprocessing the molten iron to reduce the carbon content. Several other metals are then added to givephysical characteristics to the iron. This process is used in manufacturing high grade alloy steel becauseof the improved control.

(C) Fabrication Process concerning Rolling of Steel: In this process, liquid steel is cast into ingots. Theserectangular pyramids are then rolled into blooms which square section lengths of steel. Blooms arefurther heated and rolled into billets of smaller cross-section, which are then converted into variousfabricated lengths like beams, angles, rods, flats, sheets, plates, etc.

(D) Fabrication of Steel: The fabrication requirements are designed based and depends upon thecomplexity of the process. It is done when steel materials are to be shaped in different forms.

Metal Working Processes:

Metal working processes involve operations on metals like cast iron, steel, brass, bronze, aluminium, etc,with the help of machine tools like lathe, shaper, miller, grinder, planner, etc.

The following are the metal working processes:

• Finishing, plating, honing, galvanizing, anodising.

• Forming, casting, forging, hot-rolling, extruding.

• Heat treatment, hardening, tempering, annealing, normalising.

• Joining, welding, soldering, brazing, riveting.

• Machining, milling, plaining, shaping, grinding.

Metal working processes can be described as under:

Casting: A casting may be defined as a molten material that has been poured into a prepared cavity andallowed to solidify. Casting may be classified as:

Sand Casting: It is used mainly for steel and iron and it can also be used for brass, aluminium, bronze,copper etc. and relatively large amount of metal is to be removed.


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Features of Sand Casting:

• Clay and water act as a bond.

• Compressed moist sand is used.

• Less costly.

• More machining of finished goods is required.

• More weight.

There are two methods of sand casting:

Green Sand Moulding and Dry Sand Moulding.

Green sand Moulding utilises a mould made of compressed moist sand. This method is not suitable forlarge and heavy castings. In case of Dry sand Casting, the mould surfaces are given refractory coating andare dried before the mould is closed for pouring. This method is useful for large and heavy castings.

Centrifugal Casting: The molten metal is poured into a hollow cylindrical mould, which is spinning andthe centrifugal force causes the liquid metal to flow to the outside of the mould and to remain there. Thenit is allowed to cool.

Die Casting: The dies are expensive but the advantage is that no finish machining is required in manycases. This is limited to low melting point alloys.

Investment Casting: Machining the part made out of an alloy is very difficult and in such cases, theinvestment casting is used.

Permanent Mould Casting: This is developed to avoid the above disadvantage. The advantages are smootherfinish, higher mechanical properties, good dimensional uniformity and ease of adaptability to automatichigh production. But this is having high initial cost of tooling and the size of casting is also limited by themould making equipment.

Plaster Mould Casting: Only one casting is made and then mould is destroyed. The advantages of plasterare the superior surface finish, improved metal characteristics and good dimensional accuracy. But thedisadvantage is that the mould is destroyed each time.

Forming: Forming processes are those which accomplish the rough sizing or shaping of manufacturedarticles. Any cutting tool which produces a desired contour on the work piece comes under forming process.

Heat Treatment: It is a process of heating and cooling metals in order to obtain certain desired properties.Then the right combination of hardness and toughness is achieved for enabling the cutting tools to be ableto successfully machine other metals. In addition to hardness and toughness, this process will improveheat and corrosion resistance and relieve stresses. The various types of heat treatment processes are brieflydiscussed below:

• Annealing: It refers to the heating and cooling operations which are usually applied to induce softening.Annealing softens overly hard, overly brittle part, through heating, short of critical point, followed bygradual cooling. Thus it relieves further the brittleness introduced by “hardening” and reduces metalelectricity slightly.

• Case Hardening: This process involves two operations. The first is a carburizing process where carbonis added to the outer surface by heating low carbon steel. The second operation is heat treatment ofthe carburized parts so that the outer surface becomes hard.


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• Hardening Steel: Hardening is the process of heating a metal to the decalescence point, soaking it fora considerable time to allow thorough penetration of the heat in the metal structure and then suddenlyquenching it in cold liquid.

• Normalising: Normalising is the process, in which parts are allowed to cool in still air at roomtemperature.

• Quenching: Quenching is the process of rapidly cooling the metal from a high temperature. Rapidcooling is required for obtaining high strength and hardness and the purpose is to harden steel so thatit can withstand wear and tear.

• Tempering: Heat treatment should relieve stress as stated above. But after quenching, the metal ishard and brittle. It requires reheating. Tempering is the process of reheating which will reduce thebrittleness and soften the steel.

Welding Process:

Welding is a process of joining similar metals by the application of heat, with or without application ofpressure and addition of filler material.

“Welding consists of fusing metals together while they are in the plastic or molten state.” Gas and Electricarc are used for the purpose of heat. Electrodes are used for filling in and reinforcement. There are 8welding processes as mentioned below:

• Arc welding

• Brazing

• Flow welding

• Forge welding

• Gas welding

• Induction welding

• Resistance welding

• Thermit welding

Surface Treatment:

Surface-treating processes are chemical or mechanical in nature and alter the surface characteristics of themetal.

Eleven Types of surface treatment are briefly described below:

• Anodizing: It is an electro-chemical process which gives a slight anticorrosion protection and improvesthe appearance of the product. Unlike other types of surface treatment, this does not increase thedimensions of the object. This gives a better finish. It is used in case of utensils, household appliancesetc.

• Enamelling: It is a process that bakes on a white, brittle protection finish.

• Galvanising: It is a hot-dip process which provides an anti-rust zinc coating. Zinc is used for thispurpose. This makes the surface of the metal anti-corrosive and gives a better finish.


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• Honing: It is an abrading operation for hole finishing done by a tool fitted with a bonded abrasivestone.

• Lapping: It is a surface smoothing operation by hand or machine.

• Painting: It will yield protective coating and better appearance.

• Plastic coating: This process is more durable than painting.

• Plating: It is an electrolytic process. This can be done with the help of silver, chromium, cadmium,nickel and copper. Plating provides anticorrosive finish.

• Shot Blasting: Small shots usually in the form of iron balls are made to blast on the metal required formaking it resistance to wear and tear, remove the stresses and hardness.

• Shot penning: It is air blasting the small shots against the metal surface in order to increase the hardnessof surface.

• Tumbling: Castings are tumbled in a tumbling barrel together with an abrasive substance. The frictioncreated in this way will clean the surface.

Other surface treating processes may be such as:

• Buffing etc.

• Polishing,

• Power brushing,

• Sandblasting,

• Washing,

• Waxing

1.1 Machining Process

Machinability is described as the case with which metal can be removed. The various machining processesare briefly described below:

• Blanking: In this process also “punch and die” is used for making domestic utensils or machinecomponents. It produces necessary depressions and cut-outs.

• Boring: In this case, the work piece is held by a fixture and a rotary cutter is used for boring a hole ina cylindrical shape. The existing hole may be enlarged. The boring may be straight or tapered.

• Broaching: The broach is made up of a bar of suitable length with a series of cutting edges on itssurface. These cutting edges are arranged progressively higher from the starting to the end and henceeach successive tooth removes an additional amount of material. This is done with the help of a cutterknown as “broacher”. This process also produces slots and gear teeth.

• Die Cutting: It is of two types ‘punching’ and ‘Blanking’ type. In the punching die, the metal removedby the die is a scrap, leaving a hole in the work piece. The blanking die shears through the work pieceand the metal removed by the die is the finished work piece.


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• Drilling: The work piece is clamped on the work table and drill is fed into for generating a hole andenlarging an existing hole. It is the operation of making a uniform hole through the work piece withthe help of a cutter known as ‘drill bit’. The hole can be made by rotating the job also.

• Facing: Facing is an operation of producing a smooth flat surface on the face of the job.

• Flame cutting or gas cutting: It employs an oxyacetylene torch, which may be directed either by handor pantograph and cuts up to 40inch.thickness. Flame cutting is an inexpensive method. Gas cuttingis the process of heating the metal to red hot temperature and then oxidising it by an oxygen jet. Themetal oxide is removed by the pressure of the oxygen jet. Oxyacetylene flame is used for heating. A jetof high pressure oxygen is used to effect the cutting.

• Grinding: It is a process of removing the metal form and finishing the surface of a metal object by agrinder with the help of abrasive wheel rotating at high speed. It produces smooth and accuratesurface finely finished at a quicker rate. It is suitable for repetitive jobs and for mass production.Grinding may be done on the job of any size or shape and on internal as well as external surfaces. Itensures high dimensional accuracy and surface finish. Normally the grinding follows some othermachining like turning and milling etc.

• Milling: Milling is a method of removing metal with a milling cutter in order to produce specificshape. The function of milling is to dress a flat or other surface to a fine finish, surface forming of gearteeth, keyways etc.

• Parting: In this process, a parting tool or hack-saw is used for cutting off a part from the job.

• Plaining and Shaping: In shaping, the tool is reciprocated while the work piece moves straight forwardat each stroke. In Plaining, the work piece reciprocates while the tool is moved for each new stroke.The shaper takes small work pieces and the planer takes larger work pieces.

• Punching: It produces cut-outs or perforations on plate surface e.g. holes in washer and slots in channelswith the help of combination tool known as “punch and die”.

• Reaming: The main purpose of reaming is to smooth the hole. In this process, a multiple point cuttingtool called ‘Reamer’ is used.

• Sawing: Sawing is a cutting operation done by saw. Common types are Hacksaw, Circular saw andBand saw.

• Screw cutting: It is a process of cutting screw threads on the outside surface of a job with the help ofa multiple point cutting tool known as ‘Die’.

• Slotting: This process is used to produce the slots on the job.

• Taper Turning: After turning, it produces conical shape instead of cylindrical shape.

• Tapping: In this process, a multiple cutting tool called “Tap” is used for making threads on the insidesurface of a hole.

• Turning: The work piece mounted on the lathe rotates at high speed above its axis. The path of thetool is parallel to the axis and the cutter is applied to the work piece as it turns thus chipping away themetal. Then a cylindrical external surface is formed. It is the operation of cutting and removing materialfrom the surface of the job.


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1.2 Class of Machines

Machine Tool: It is a power driven mechanical appliance or device used for removing metal from surfacesof jobs to give them the proper shape, size and other characteristics.

Classes of Machine Tools:

Machine Tools are now so numerous in type that they do not admit of any fixed classification. One possibleclassification is under two groups, namely

(1) General purpose or Standard Machine Tools, (2) Special Purpose Machine Tools.

The General Purpose or Standard Machine Tools form the most essential part of the equipment of aworkshop. They are called Standard Machine Tools in the sense that they are very common and essentialand a list of most important Machine Tools used in a workshop will invariably include these machines.

Many of them are suitable for many different types of operations. A lathe, which is the most importantmachine belonging to this group, can be employed for the general purposes of turning (producing cylindricalsurfaces on a job), drilling (making a hole), screw cutting, producing flat surfaces, cutting teeth on a gear,cutting cylindrical holes in job, etc.

A special purpose Machine Tool performs only a limited number of specialised operations with greatspeed and precision. It does the same job as can be done by some General Purpose Machine, but it supersedesthe latter on account of its high speed and accuracy. As for example, boring operation can be done on alathe but the same operation can be done with greater speed and efficiency on a Boring Machine which hasbeen developed for doing mainly the operation of boring.

The General Purpose Machine has great adaptability, but on account of the present day need for high-speed production of parts of great accuracy, Special Purpose Machines have been developed by sacrificingadaptability to specialisation.

Multi-purpose and Single-purpose Machine Tools:

Machine Tools are sometimes classified under Multi-purpose and Single-purpose machines.

A Multi-purpose Machine Tool is capable of doing a number of different types of operation. A CentreLathe, a Milling Machine, a Grinding Machine, etc. are examples, of Multi-purpose Machines.

A Single-purpose Machine is one that is capable of doing only one particular kind of operation, but is notrestricted to one particular job. As for example, a Shell Turning Lathe is designed specially for shell turningand is not adopted for general Lathe work. A Hobbing Machine, a Honing Machine, etc. are other examplesof Single-purpose Machines.

Although a Single-purpose Machine has restricted field of use, it can do the particular operation for whichit is employed at great speed and with a high degree of accuracy. When a large number of interchangeableparts are to be quickly produced, Special-purpose Machines are of great importance.

General Purpose or Standard Machine Tools:

The following are some General Purpose Machine Tools employed to do various operations:

• Drilling Machines- These Machines are used for drilling holes of different diameters in vertical orinclined directions.


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• Grinding Machines - Grinding Machines are used for machining parts having cylindrical and taperedsurfaces, for grinding internal surfaces and flat surfaces and for grinding slender bars and delicateparts.

• Lathes (Centre or Engine Lathe) - The operations performed on Lathes normally are Turning, Boringand Screw-cutting. Over and above these normal operations a Lathe can be adapted for the operationsof Milling, Drilling, Slotting, Grinding, Gear and Rack Cutting, Lapping, Facing, Reaming, Polishing,Knurling, Tapping and Parting off.

• Milling Machines - These Machines are employed for producing plain and irregular surfaces, cuttingslots by using special tools, cutting threads on dies, worms and screws and for gear-cutting.

• Plaining Machines - These Machines are used for machining flat surfaces and cutting Key-ways onlarger jobs.

• Shaping Machines - The operations performed on Shaping Machines are machining of flat surfaces,cutting of Key-ways and T-slots on jobs not exceeding 12 inches in length.

1.3 Special Purpose Machines

Honing Machine: Honing, lapping and super-finishing equipment are used to improve surface finish orgeometry to tight tolerances. Honing machines use small, bonded abrasive stones or super abrasive sticksmounted in a fixture that rotates and reciprocates when applied to the surface or bore being finished.Honing is used to correct the geometry and alignment of holes and to produce the surface required for theapplication. Honing is final finishing operation conducted on a surface, typically of an inside cylinder(bore), for example automobile engine cylinder. Abrasive stones are used to remove minute amounts ofmaterial in order to tighten the tolerance on cylindricity.

Honing is a surface finish operation, not a gross geometry-modifying operation. Honing Holders apply aslight, uniform, pressure using a number of abrasive sticks (Honing Sticks) that wipe over the entire surfaceto be honed thus removing material from the surface. Honing is the final process used in the machiningcylinder bores, either during manufacture or in re-sizing (re-boring). Honing is used as a process to bothremove the final amount of metal to get a cylinder bore to within the required size limits, and to put asurface on the cylinder bore which will give good life span, and aid lubrication and oil consumptioncharacteristics in use.

Honing is not only carried out on cylinder bores for internal combustion engines but oncompressors, hydraulics components and probably dozens of other applications.

Other Special Purpose Machine Tools:

The following list includes some of the most important Special Purpose Machines Tools:

• Boring Machines - These machines are used mainly for boring operations. By using Special Tools theymay be adapted to facing the end of work and cutting internal thread.

• Broaching Machines - These machines are used for altering the size of finish of holes in metallic parts.

• Gear-Cutting Machines - These machines are used for cutting gears.

• Honing Machines - (already described above under separate head) : These machines are used foreliminating local irregularities of surface and giving good finishing.


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• Lapping Machines - These machines are used for finishing cuts.

• Turret and Capstan Lathes - The operations performed on these machines by using appropriate toolsare Turning, Forming, Parting off, Boring and Screw-cutting, the operations can be automaticallyperformed in any sequence.

Machine Tools used in a Workshop:

A workshop should, in the first place, contain all the General Purpose Machines and some Special PurposeMachines according to need. A special Purpose Machine should be installed if it can be engaged for areasonable percentage of its useful life. Otherwise General Purpose Machines should be employed, so thatidle capacities of the machines may be reduced to the minimum.

The following machines are most usually found in an up-to-date machine shop manufacturing lightmachines and their interchangeable parts:

(1) Lathes – (a) Centre or Engine Lathe, (b) Capstan Lathe, and (c) Turret-Lathe.

Lathes are used mainly for Turning, Boring and Screw-cutting and can be adapted for the operations ofMilling, Drilling, Slotting, Grinding, Gear and Rack cutting, Lapping, Facing, Reaming, Polishing, andKnurling, Tapping and Parting off.

The Lathe:

The following are the different kinds of Lathes : (1) Centre or Engine Lathe, (2) Capstan Lathe, (3) TurretLathe and (4) Automatic Lathe.

Of these the Centre or Engine Lathe has been classified as a General Purpose Machine and the others asSpecial Purpose Machines. The Centre Lathes are extremely versatile and with proper equipment can doan endless variety of work. Capstan, Turret and Automatic Lathes have limited scope of work but can dospecialized jobs with great speed and accuracy. With the modern tendency towards specialisation, CentreLathe has been to a large extent superseded by Capstan, Turret and Automatic Lathes and Grinders.

By the term “Engine Lathe” is meant the Ordinary Centre Lathe with at least one automatic feed, either forthe “traverse”, or for the “surfacing”.

When both the traverse and the surfacing motions are power operating the Lathe is called a “Self-acting,Sliding and Surfacing Lathe”.

When a Self-acting, Sliding and Surfacing Lathe is provided with a Lead Screw for Screw-cutting, it iscalled a “Self-acting, Sliding, Surfacing and Screw-cutting Lathe”.

The Principle of Operation of the Lathe:

In a Lathe a great deal of work is in the form of cylindrical bars. To product cylindrical surfaces the job issupported on two parts known as the Headstock and the Tailstock and is uniformly rotated by means of amechanical device. It is clear that if a cutting tool is applied on the rotating job, cylindrical surfaces ofspecific diameter can be produced. The cutting tool can be operated on the entire surface of the job bymoving the former parallel to the axis of rotation of the job. The rotation of the job and the motion of thecutting tool are produced by power.

Principal Parts of a Lathe:

(1) The Bed, (2) The Headstock, (3) The Tailstock, which is also called the Loose Headstock,(4) The Carriag, (5) The Gearbox, (6) The Feed Shaft and the Lead Screw.


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Figure of Lathe

Lathe Attachment :

Lathe Attachments are employed for increasing the adaptability of the Lathe for handling special kinds ofjobs. The attachments are either devices for holding jobs or are Miscellaneous Tools suitable for differentoperations.

The following devices are meant to hold jobs:

(1) Special Centres and Shaft Supports.

• When the usual Headstock and Tailstock Centres are unsuitable for specialkind of work, Special Centres are used.

• When a long slender shaft is turned between centres if is supportedby a de-vice called Travelling Stay or Steady Rest attached to the Lathe Spindle.

(2) Mandrel,

(3) Chuck,

(4) Faceplate.

Miscellaneous Tools : For Drilling and Boring, Special Tools are attached to the Tailstock, so that when thejob rotates, a hole is made into the job. Suitable Tools are used for other purposes.

Operations done on a Centre Lathe:

The following are the operations usually done on a Center Lathe: –

• Boring : Boring does the operation of enlarging the diameter of a hole already exist in a job. SmallBoring operations are often done in the Lathe either as a single operation or preparatory to Threading,Reamering or other operations.

• Screw Cutting: this is the operation of cutting screw threads on the external surface of a cylindricaljob.

• Turning: Turning means producing cylindrical surface on a job. In this operation material from the


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surface of the job is removed by rotating the job against a Tool point, so as to produce a cylindricalsurface. The diameter of the work may be ascertained by Callipers, Micrometers or fixed gauges.

In additional to these normal uses, a Centre Lathe can readily be adapted for the following operations: –

• Milling: Milling is the process of giving a specific shape or form to a job by cutting it by means of arevolving multiple point tool.

• Drilling: Drilling is the operation of making a hole in a job. The Drill is inserted into the TailstockSpindle and the Tailstock is clamped to the job in the necessary position. The feed is affected by theTailstock hand wheel.

• Slotting: This is the operation of making a long narrow aperture for something to be inserted orworked in.

• Grinding: The Grinding operation consists in applying a wheel of abrasive character rotating at highspeed to a metallic surface for causing it to wear.

• Gear and Rack cutting: A Gear is a toothed wheel which, having the teeth meshed with those ofanother Gear, can cause one shaft to drive another. A gear having the teeth in a straight line, insteadof on a circle, is called a Rack.

• Polishing: Polishing processes on job surfaces may be undertaken on a Centre Lathe.

• Lapping: Lapping is the process of polishing a work by-means of abrasive materials, to give finefinish. Lapping operation may be undertaken after grinding to give fine adjustments.

• Facing: this is an operation of producing a flat surface and removing material from the surface. Theoperation consists in rotating the job on an axis of rotation perpendicular to the flat surface.

• Reaming: The Reaming operation consists in (1) enlarging existing drilled holes; (2) making, a parallelhole into a tapered hole and bringing existing holes actually to size.

• Knurling: Knurling is the process of producing a series of; right-hand and left-hand fine grooves onthe surface of a job to facilitate handing or to secure a better hold. Knurling operation may beaccomplished by the use of either a single or double wheel Knurling Tool. The work must be mountedrigidly; either well up against Chuck Jaws or, in case of a long, slender work, a suitable Steady mustbe employed.

• Tapping: Tapping is the operation of forming threads on the interior surface of a job. For tappingthreads a Tap should be used and the job should be very slowly rotated by engaging the Back Gearsystem.

• Parting Off: when work is machined from bar, the completed job can be parted off from the bar on theCentre Lathe.

Special purpose Lapping Machines are also used for this purpose.

Capstan and Turret Lathes:

The Centre Lathe is a versatile machine and can be used for many different kinds of operations. It howeverrequires for its operation a skilled operator whose hands produce intricate and precise work. With thedemand for speed and accuracy accompanied with low cost the need arose for more rapid results from lessskilled personnel. As a result, Capstan and Turret Lathes were developed.


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Capstan Lathe

The Capstan Lathe has the following principal parts

(1) A Bed.

(2) An all-geared Head-stock.

(3) A Saddle, in which the Cross-slide carries a revolving square Turret that can hold four Tools at once.The Turret is unlocked by means of a lever, and by turning it manually, any of the Tools in the Turretcan be brought into the working position.

(4) The main Turret which replaces the Tailstock of the Centre Lathe.

Operations performed on a Capstan Lathe:

The following are the important operations performed on the Capstan Lathe:—

1. Turning, that is producing cylindrical surfaces.

2. Forming, that is turning intricate shapes by one traverse of a Tool, instead of several movements.

3. Parting-off, that is cutting away of a completed job from a bar.

4. Boring, that is enlarging the diameter of a hole already existing in a job.

5. Screw cutting, which is cutting screw threads on the external surface of a cylindrical job.

Turret Lathes

Turret Lathe is almost similar in construction to Capstan Lathe. Its principal parts are the following:—

(1) A bed.

(2) An alt-geared Headstock.

(3) A Saddle in which a revolving square Turret is carried by the Cross-Slide.

(4) The main hexagonal Turret which carries six different Tools in special holders that are bolted up tothe faces.

Operations performed on a Turret Lathe:

The operations performed on a Turret Lathe and on a Capstan Lathe are the same. Generally, however,Turret Lathes are suitable for machining bigger part than Capstan Lathes.

The following are the important operations performed on a Turret Lathe:

(1) Turning,

(2) Forming,

(3) Parting-off,

(4) Boring,

(5) Screw-cutting.

Plaining Machines: are used for machining flat surfaces and cutting key-ways on large jobs. The functionof this machine is same as the shaping machine but here the tool is held fixed and gives the required feed


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whereas the work piece is rigidly fitted on the work table with a reciprocating motion. The length of theworkpiece is more than 1 metre and are normally massive.

Figure of Plaining Machines

The main parts of the machine are bed, work table, cross slide pillars and tool holders as shown in figure.Tool holders, (for at least two tools) are mounted on cross beam and have provision to slide up or downalong two vertical pillars fixed on two sides of the bed. After first setting, this cross beam is firmly clampedand then during operation the clapper box containing tool holders slide to give the feed.

Slotting Machines: are used for machining vertical or inclined flat surfaces, making holes with sharp cornersand tapered holes and for cutting T-slots and V-ways.

For mass production, Single-purpose Machines may be used in addition to the machines mentioned above.

This is another type of a machine to produce machined fiat surface like shaper and planer. The differencewith shaper or planer is that slotter makes vertical surface machining whereas the others produce horizontalmachining. As such it is sometime called as “vertical shaper”, since the ram holding the tool moves up anddown. Base and pillar is a composite L-shaped cast structure, the pillar having a vertical machined way forthe reciprocating movement of the ram. The work piece is held firm on a round table with saddle which isfitted on a horizontal bed. The round table is suitable for adjustment for the movements at right angles inhorizontal plane and can also rotate about the vertical axis. The movement of the job gives the desired feedwhereas movement of ram produces cutting.

The slotting machine is suitable for making keyways on inside surface of the bore of a pulley or gear or onthe outside surface of a round. A typical shaping machine is shown in figure.


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Figure of Slotting Machine

Shaping Machine:

Shaping, like Plaining, is the machining of flat metallic surfaces by single-point cutting tools. Shaping iscarried out on small areas of metal not usually exceeding one foot in length. It consists of a massive bodysupporting the Ram which slides along ways provided on the upper surface.

There is a Saddle which supports the Work Table. The Saddle can slide up or down the face of the bodyand adjusts the height of the Work Table to accommodate jobs of different heights. When shapingcommences, the Saddle is locked, so as to keep the job at a constant height.

Figure of Shaping Machine

Uses of a Shaping Machine:

A Shaping Machine has the following uses:

1. Machining flat surfaces: The areas to be machined are how ever small, such as small Bed Plates ofMachines.


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2. Cutting Key-ways: For this purpose an aperture in the body of the machine enables shafts to be runparallel to the Ram for the provision of the Keyways.

3. Cutting T-slats: T-steps are cut in three operations. Firstly, a plain slot is cut to the required depth.Then each side is undercut by means of suitable undercutting tool to cut the side slots so that finallya slot is produced resembling the letter ‘T’.

Drilling Machines:

Drilling Machines may be broadly grouped into the following four classes, such as

1. Sensitive Drilling Machines,

2. Pillar Drilling Machines,

3. Radial Drilling Machines,

4. Multiple Spindle Drilling Machines.

Sensitive Drilling Machines: Sensitive Drilling Machines are designed to take Drills up to about 8/l6 inch indiameter. They are provided with hand feed for driving the Drill forward.

The Sensitive Drills are used for making small holes only. The hand feed is provided by hand by movingthe drill spindle by means of a rack and pinion arrangement. In operating these drills, pressure applied onthe drill bit can be felt by the operator by his senses. These Drills are therefore called Sensitive Drills.

Pillar Drilling Machine:

A Pillar Drilling Machine is an upright Drilling Machine as distinct from a Radial Drilling Machine Itconsists of a Work Table on which the job to be drilled is placed.

The Work Table can be adjusted sideways and vertically to accommodate jobs of different heights.

Figure of Pillar Drilling Machine


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The arm carries the Spindle in a Saddle which accommodates the Spindle driving motor and the Gearboxesfor providing different Spindle speeds and feeds.

Radial Drilling Machine:

Radial Drilling Machine which has a horizontal arm which is attached to a sleeve capable of moving upand down a vertical pillar and can be clamped at any position. The sleeve can freely rotate about the pillarwhich is fixed to the base.

Figure of Milling Machine

Figure of Radial Drilling Machine

The arm carries the spindle in a saddle which accommodates the spindle driving motor and the gearboxesfor providing different spindle speeds and feeds.

Milling Machine:

Milling is the operation of removing metal by means of a rotating cutter against which the work is fed. Themilling cutters are generally discs of cylindrical Tools-usually made of high-speed steel and have serrationsor teeth about their edges.

The face of the tooth is given a definite cutting rake or angle appropriate for the material of the job to bemachined.

A Milling Machine is a very important machine tool in a modern workshop. The general purpose MillingMachine may be broadly classified under (a) the horizontal milling machine and (b) the vertical milling machine.


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In a Vertical Milling Machine the working spindle is vertical, whereas in a Horizontal Milling Machine it ishorizontal.

A vertical Milling Machine is used in milling inside a recess or cavity and in performing operations whichcannot be done on a Horizontal Milling Machine. There are many operations which can foe performed onthe horizontal as well as on the Vertical Milling Machines.

Universal Milling Machine:

In a Universal Milling Machine the work-table can be swivelled about a vertical axis, so that its direction oftravel may be at right angles or at any other angles to the spindle axis. This enables helical grooves to be cuton the outside of a cylindrical work, for instance cutting of helical teeth on gear blanks, milling of helicalgroove on a twist drill, etc. These operations cannot be performed on a plain milling machine in which thetravel of the work-table is set permanently at right angles to the spindle axis.

Uses of a Milling Machine:

The following are the uses of a Milling Machine:—

1. Forming plain and irregular surfaces.

2. Cutting slots by using suitable Tools.

3. Cutting threads on dies, worms, screw, etc. internally or externally. Cutting threads on the MillingMachine is known as Thread Milling.

4. Gear cutting by the uses of Special Gear Cutters.

Grinding Machines are used for machining cylindrical and tapered surfaces, for grinding internal and flatsurfaces and slender bars and delicate parts.

The Special Grinding:

The Grinding Process consists in machining jobs by means of abrasive wheels. In an abrasive wheel a largenumber of abrasive particles are held together at the periphery by means of a, bonding material. Theabrasive particles act like minute cutters. When the abrasive wheel is rotated against a job, fine surface isproduced by the wearing away of the materials on the job. The chips produced are usually so minute thatthey cannot be seen with the naked eye. The abrasive particles may be Aluminium Oxide, Silicon Carbideor Diamonds in the form of “dust”.

Main Types of Grinders and their Uses:

There are four main types of Grinding Machines in common use:

1. The Cylindrical Grinders,

2. The Internal Grinding Machines,

3. The Flat Surface Grinding Machines,

4. The Centreless Grinding Machines.

The Cylindrical Grinding Machines are subdivided into. (1) Plan Grinders, (2) Universal Grinders.

The Cylindrical Grinders are used for machining parts like shafts, spindles, rollers, etc., having cylindricalsurfaces and also for machining tapered parts, cams, eccentrics, soulders of shafts, etc. The Internal Grinding


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Machines grind the internal surfaces of cylinders and other parts. Flat Surface Grinders are designed tomachine every type of flat surface. A Centreless Grinder produces more accurately a cylindrical surface.

Of the Cylindrical Grinders the Plain Grinder is used for machine parts either cylindrical or tapered infrom The Universal Grinder has a wider range of application and can be used for grinding work at anydesired angle. It can be adapted to Internal Grinding and is useful in the production of part used in smallmachines and special tools.

Figure of Grinding Machine

The Hobbing Machine:

The Hobbing Machine is a Single Purpose Machine Tool designed for cutting gears. Hobbing operation isthe most accurate way of cutting Gears.

The cutting tool used in the Hobbing Machine is called the Hob which is in the form of a worm. A Hob isnothing more than a screw provided with “gashes” or “flutes” forming cutting edges. The cutting action isprovided by the rotation of the hob, in conjunction with the gashing that provides the cutting edges.

The Hobbing Machine may be used to generate Spur Gears or Helical Gears. The Spur Gears have theirteeth parallel to the axis of rotation of the Gear, while in Helical Gears the teeth are not parallel to the axis.

Figure of Hobbing MachineProblem 1 :A shaft 1,000 mm. in length is being machined on a lathe. If the spindle executes 500 r.p.m. and the feed is0.20 mm. per revolution, how long will it take the cutter to pass down the entire length of the shaft?

Solution:

Since the feed is 0.20 mm. per revolution, the number of revolutions in passing

1,000 m.m.=1,000

0.20 m.m. =5,000. Since the spindle executes 500 r.p.m.,

the Time required =5,000 500

m.m. = 10 min.


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Problem: 2

Find the machining cost of a M. S. bar on a lathe from the following data: R. P. M. of the Job = 500. Feed oftool per revolution of job = 0.5 mm. Depth of cut = 2 mm. Diameter of raw material = 60 mm. Diameter of

finished job = 40 mm. Length of Job = 1000 mm. Machining cost = Rs. 3 per hour.

Solution :

Number of revolutions in one traverse of 1000 mm.=1,000 mm.0.5 mm.

= 10,000

5 = 2000

Total depth of cut = ½ (60-40) mm. = 10mm. Cross-feed = 2 mm.

The number of transverse traverse over the job from end to end = 10/2 times = 5 times.

Thus the total number of revolutions of the job = 2,000 × 5 = 10,000 R. P. M. of the job = 500

Machine time =10,000

500 min. = 20 min. = 1/3hr.

Cost of machining = Rs. 3×1/3 = Re.1.

Problem: 3

A shaft 500 mm. in diameter and 1 metre long is to be turned at a speed of 280 r.p.m. If the feed is 0.25 mm.per revolution, calculate the time taken for one pass of the cutter.

Solution:

Length of the shaft = l metre = 1,000 mm.

Number of revolutions =1,000 mm.0.25 mm.

=1,00,000

25= 4,000

Time required for one pass of the cutter = 4,000280

min. = 1007

min. = 14.29 min.

Jigs and Fixtures:

A Jig is an appliance to which a work to be machined can be, fastened and which contains a device forguiding the tool, so that the tool and the work are accurately located with respect to each other. A Fixtureis an appliance which holds the work when it is machined.

Uses of Jigs and Fixtures:

(1) Jigs quickly and accurately guide the tools. They render difficult operations easier, speedier and yetmore accurate.

(2) Jigs are suitable is mass production for producing accurately machined interchangeable parts.

(3) Fixtures are essential almost all machine work, because the work must be firmly held when the toolswork.

(4) Fixtures used in conjunction with jigs increase speed and accuracy of work.


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1.4 Pumps, Motors, Transformers, Electrical Drivers

Pumps

A pump is a device used to move liquids. A pump moves liquids from lower pressure to higher pressure,and overcomes this difference in pressure by adding energy to the system (such as a water system). A gaspump is generally called a compressor, except in very low pressure-rise applications, such as in heating,ventilating, and air-conditioning, where the operative equipment consists of fans or blowers.

Pumps work by using mechanical forces to push the material, either by physically lifting, or by the force ofcompression.

Pumps fall into two major groups: Rotodynamic pumps and positive displacement pumps. Their namesdescribe the method for moving a fluid. Rotodynamic pumps are based on bladed impellers which rotatewithin the fluid to impart a tangential acceleration to the fluid and a consequent increase in the energy ofthe fluid. The purpose of the pump is to convert this energy into pressure energy of the fluid to be used inthe associated piping system. A positive displacement pump causes a liquid or gas to move by trapping afixed amount of fluid or gas and then forcing (displacing) that trapped volume into the discharge pipe.Positive displacement pumps can be further classified as either rotary-type (for example the rotary vane)or lobe pumps similar to oil pumps used in car engines. Another common type is the helical twisted Rootspump. The low pulsation rate and gentle performance of this Roots-type positive displacement pump isachieved due to a combination of its two 90° helical twisted rotors, and a triangular shaped sealing lineconfiguration, both at the point of suction and at the point of discharge. This design produces a continuousand non-verticals flow with equal volume. High capacity industrial “air compressors” have been designedto employ this principle as well as most “superchargers” used on internal combustion engines.

Reciprocating-type pumps use a piston and cylinder arrangement with suction and discharge valvesintegrated into the pump. Pumps in this category range from having “simplex” one cylinder; to in somecases “quad” four cylinders or more. Most reciprocating-type pumps are “duplex” (two) or “triplex” (three)cylinder. Furthermore, they are either “single acting” independent suction and discharge strokes or “doubleacting” suction and discharge in both directions. The pumps can be powered by air, steam or through abelt drive from an engine or motor. This type of pump was used extensively in the early days of steampropulsion (19th century) as boiler feed water pumps. Though still used today, reciprocating pumps aretypically used for pumping highly viscous fluids including concrete and heavy oils.

Another modern application of positive displacement pumps are compressed air-powered double-diaphragm pumps, commonly called Sandpiper or Wilden Pumps after their major manufacturers. Theyare relatively inexpensive, and are used extensively for pumping water out of bunds, or pumping lowvolumes of reactants out of storage drums.

Centrifugal pumps are Rotodynamic pumps which convert Mechanical energy into Hydraulic energy bycentripetal force on the liquid. Typically, a rotating impeller increases the velocity of the fluid. The casing,or volute, of the pump then acts to convert this increased velocity into an increase in pressure. So if themechanical energy is converted into a pressure head by centripetal force, the pump is classified as centrifugal.Such pumps are found in virtually every industry, and in domestic service in developed countries forwashing machines, dishwashers, swimming pools, and water supply.

After motors, centrifugal pumps are arguably the most common machine, and they are a significant user


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of energy. Given design margins, it is not unusual for a pump to be found to be over-sized, having beenselected poorly for its intended duty. Running a constant speed pump throttled causes energy waste. Acondition monitoring test can detect this condition and help size a smaller impeller, either new, or bymachining the initial one, to achieve great energy reduction.

Pumps also wear internally, at a rate varying with the liquid pumped, materials of construction and operatingregime. Again, condition monitoring can be applied to detect and quantify the extent and rate of wear andalso help decide when overhaul is justified on an energy-saving basis.

Kinetic Pumps: The features are:-

• Continuous energy addition

• Conversion of added energy to increase in kinetic energy (increase in velocity)

• Conversion increased velocity to increase in pressure

• Conversion of Kinetic head to Pressure Head

• Meet all heads like Kinetic, Potential, and Pressure

Motors:

An electric motor uses electrical energy to produce mechanical energy. The reverse process that of usingmechanical energy to produce electrical energy is accomplished by a generator or dynamo. Traction motorsused on locomotives often perform both tasks if the locomotive is equipped with dynamic brakes. Electricmotors are found in household appliances such as fans, refrigerators, washing machines, pool pumps,floor vacuums, and fan-forced ovens. Most electric motors work by electromagnetism.

The classic division of electric motors has been that of DC types and AC types.

Transformers:

A transformer is a device that transfers electrical energy from one circuit to another through inductivelycoupled wires. A changing current in the first circuit (the primary) creates a changing magnetic field; inturn, this magnetic field induces a changing voltage in the second circuit (the secondary). By adding a loadto the secondary circuit, one can make current flow in the transformer, thus transferring energy from onecircuit to the other.

A key application of transformers is to reduce the current before transmitting electrical energy over longdistances through wires. Most wires have resistance and so dissipate electrical energy at a rate proportionalto the square of the current through the wire. By transforming electrical power to a high-voltage, andtherefore low-current form for transmission and back again afterwards, transformers enable the economictransmission of power over long distances. Consequently, transformers have shaped the electricity supplyindustry, permitting generation to be located remotely from points of demand. Transformers are some ofthe most efficient electrical ‘machines’, with some large units able to transfer 99.75% of their input powerto their output.


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Figure of Transformer

It is defined as a stationary induction apparatus for transforming electrical energy in AC system from onecircuit to another by changing voltage but without changing frequency. In its simplest form it consists oftwo insulated coils wound over a common magnetic core of soft iron lamination as shown in figure. Thewinding which receives power from the source is called “primary winding” and the winding which deliverselectric power to the load is known as “secondary winding”. The main purpose of transformer is to changevoltage between the two windings.

The operating principle is very simple. When an AC is fed to primary winding, current flows in the windingwhich produces a magnetic flux within the core known as “mutual flux” which is proportional to thenumber of winding and the frequency of AC supply. This flux links up the secondary winding to producean induced emf proportional to the number of turns of secondary winding. Thus if

E1 = voltage of primary winding

N1 = number of turns in primary coil

E2 = voltage of secondary winding

N2 = number of turns in secondary coil.

Then, 1

2

EE =

1

2

NN

If E2 is greater than E1 it is called a “step up” transformer as it increases the voltage and if E2 is less than E1,

it is known as “step down” transformer. Since the ampere turns of primary and secondary winding remains

the same and if I1 and I2 be the primary and secondary currents, then,

N1I1= N2I2

1 1 2

2 2 1

∴ = =E N IE N I

or, 2

1

N

N =

1

2

I

Ior, E1I1=E2I2

Auto Transformer: For a very small variation in output and input voltage “double wound” transformer asexplained becomes expensive and “Auto wound transformer” becomes cheaper. Auto transformer has asingle winding and the secondary voltage is tapped from that winding as shown in figure.


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Electrical Drivers:

A Driver is an electronic component used to control another electronic component, such as a high powertransistor. A transistor is a semiconductor device, commonly used as an amplifier or an electrically controlledswitch. The transistor is the fundamental building block of the circuitry in computers, cellular phones, andall other modern electronic devices.

Because of its fast response and accuracy, the transistor is used in a wide variety of digital and analogfunctions, including amplification, switching, regulation, signal modulation, and oscillators.

1.5 Classification of Industries Based on Production Process

Type of Industry : The type of industry and the method of the manufacturing process exercise a significantinfluence on plant layout.

Industries in this context may be broadly classified into four types:

(a) Synthetic,

(b) Analytical,

(c) Conditioning; and

(d) Extractive.

Extractive industries involving the separation of one element from another, as in the case of metal form theore. Air conditioning industry involves a change in the physical properties. Metal working industries,foundries and leather tanning concerns condition their raw materials to have the end products. An oilrefinery, for example, yields naphtha, gasoline, paraffin, tar and kerosene. Similarly, a sugar mill gives,besides sugar, bagasse and molasses. A synthetic industry, also called the assembling industry, involvesthe production of a product by the use of various elements. In other words, various elements go into themanufacture of an end-product. The chemical industry and the paper industry are synthetic industries.Light and heavy engineering and watch-making industries, in which several components are assembled toget the final products, also fall under synthetic industries.

Each of the above types of industries may be further classified into two types, namely:

(a) Intermittent industries; and

(b) Continuous industries.


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This classification is based on the method of manufacture. The former type of industries manufacturesdifferent components on different machines and assembles them to get the end products. Continuousindustries uninterruptedly produce one or two products of a standardise nature. It is needless to emphasisethat the layout designer should keep in mind the type of industry and the method of the manufacturingprocess while Plaining a layout.

Intermittent production system situations are those where the facilities must be flexible enough to handlea wide variety of products and sizes or where the basic nature of activity imposes change of importantcharacteristics the input (change in product design). Under this system no single sequence of operations isappropriate and therefore standardized materials or machines cannot be used. Under this type ofmanufacturing, production is done in lots rather that on a continuous flow basis. It is done more often onthe basis of customer orders.

The chief characteristics of intermittent industries are that components are made for inventory but they arecombined differently for different customers. The finished product is heterogeneous but within a range ofstandardized options assembled, by the producers. Since production is partly for stock and partly forcustomer demand, there are problems to be met in scheduling, forecasting, control and coordination.Examples of such industries are automobile industry, electrical goods manufacturing plants, printing pressesetc.

Intermittent system of production may further be divided into two types, namely (a) job and (b) batchproduction.

(a) Job Production - In this system, goods are produced according to the orders of the customers.Continuous demand of such items is hot assured and therefore production is done only when theorders for the manufacturing of items are produced from the customers.

(b) Batch Production - Under this system, the manufacturing is done in batches or groups or lots eitheron the basis of customer’s order or with a hope of a continuous demand of the product. Under thissystem, medium scale production is warranted.

In batch production, machines and equipment are made available for the next batch as soon as the productionof first batch is completed.

Continuous production situations are those where facilities are standardized as to routing and flow sincethe raw material or inputs are standardized. Therefore, a standard set of processes and sequences of processesare adopted. Such processes are adopted by concerns which produce goods or services continuously byputting them through a series of successive connected operations in anticipation of customer demandrather than in response to customer orders. Examples of industries using such technology are petroleum,chemicals, steel and sugar industry.

We can again classify the continuous industries into

(1) Mass production system and (2) Process production system.

(a) Mass Production - This system of production is used by concerns where manufacturing is carried oncontinuously in anticipation of demand, though demand of the product may not be uniform throughoutthe year. Standardization is the keynote of mass production. Standardized raw materials and machinesare used to produce standardized products through standardized process of production.

(b) Process Production - This system is an extended form of mass production where production is carried


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on continuously through, a uniform predetermined sequence of operations. Generally under thissystem finished product of one process is used in the next process as a raw material till the last process.Process production calls for the setting up of automatic machines and equipment as far as possible.Large industries like petroleum refining, heavy chemical industries generally use this system ofproduction.

1.6 Technological Aspects of Different Production Process Like Power, Pollution control& Recovery Process

Power: This is an important factor in the industry and can provide the competitive edge. From the days ofa hefty power consumption of 140 units per tonne of cement, the consumption is about 90-100 units.Availability of power is another problem and almost all companies have gone in for massive powergenerating sets to improve their working.

Recovery Process - Waste Management:

There are various ways in which one could categorise waste. However, the generally accepted classificationis given below:

(i) Classification on the basis of Resources, i.e., how much of a particular resource has been wasted.

(ii) Classification on the basis of property i.e. whether the materials that have been wasted arehazardous to life and environment or whether they fall in the category of non-hazardous.

(iii) Classification on the basis of the recoverability of resources.

(iv) Classification on the basis of origin of waste, i.e., whether it is commercial waste or industrialwaste, residential waste or office waste and construction waste or agricultural waste.

The chief objectives of waste management are as follows:

• Minimisation of overall waste in any operating system under scrutiny.

• Maximisation of previous resources, so that these are not frittered away and the opportunity cost isbare minimum.

• To cut down on all the unnecessary activities which do not add value to the system.

• To increase the profitability of the operation followed by different organisations.

• To inculcate a sense of cost-effectiveness which could trigger off the prudent practices of Total CostManagement (TCM).

• To follow the ethics and the principles of Total Quality Management (TQM).

• To aspire for international recognition, vital to face the competition prevailing in the current global aswell as the domestic market.

Attributes of effective waste collection system:

An efficient waste collection system serves the organisation in more ways than one. It may be extremelycostly to install and operate, but once handled properly, provides substantial savings to the organsiationby providing the operating costs and other overheads. The structure of an effective waste collection systemdepends on the following key factors.


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1. Identify the waste.

2. Waste-separation at source.

3. Decide about the quantity to be stored in a particular container.

4. Decide about the physical dimensions other important attributes of the container.

5. Collect the waste in these containers.

6. Make projections of the rate of waste generation.

7. Depending on the above, ensure that the waste collection is timely and proper without any delays orbottlenecks.

8. Induce some kind of benefits—whether monetary inducements or otherwise to the employees collectingthe waste at source.

9. Motivational and other techniques can be effectively used to achieve the above defined result.

10. Make provisions to transport the waste so collected to the salvage industry or the localised unit, as thecase may be.

11. Collect the available organic wastes.

12. Ensure that the inorganic wastes are not left behind.

13. Clearly differentiate between the different types of waste.

14. Handling each container on the basis of the waste-type stored in these.

15. Keep an emergency or contingency plan ready.

Recycling of wastes

Often the waste generated across the industries can be recycled and used again and again. However, it isnot as simple as it appears to be. Certain appropriate recycling projects have to be created to achieve suchresults.

Features of waste disposal system:

The salient features of an effective waste proposal system are as follows:

1. Easy to install and operate.

2. Economical from the cash outflow point of view.

3. Convenient and not highly complex.

4. Within the budgetary constraints.

5. Approved by the legislation and other statutory authorities in force.

6. Flexibility and not rigidity in operations.

7. Economies of scale.

8. Does not require highly skilled labour force for its operation.

9. Adaptability in the context of the needs and objectives of the organsiation.


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Pollution Control:

Pollution can be of four types, on the basis of the forms:

(i) Solid – type , (ii) Liquid – type, (iii) Gaseous pollutants, (iv) Hybrid-type, having features andcharacteristics of one or more of the above three.

Definition and meaning of noise-pollution - Noise may be defined as an unwelcome, unpleasant, unwantedand unavoidable sound. For example it can be classified as under:

Unpleasant Noise - Sources of such noise are, predominantly:

(i) Industrial outlets.

(ii) Increased automation.

(iii) Lack of work culture.

(iv) Lack of awareness.

(v) No concern for the environmental and other allied issues.

Unavoidable Noise: Unavoidable noise is a part and parcel of our daily life. We simply cannot escape fromit. It is one of the prices one has to pay foil modern living. These are from:

(i) Regular sources.

(ii) Work-places/stations.

(iii) Use of machines.

(iv) Limited and restricted choices.

Other types could be vibration. People exposed to it for longer periods are likely to suffer from fatigue andexhaustion – which may prove to be quite costly in the long run.

Objectives of pollution control

Although the main aims of pollution control are simple enough, i.e. “to make the world a better place tolive in”, the sub-objectives and the strategies adopted for this purpose are given below:

1. Environmental preservance.

2. Ecological balance.

3. Dignity of life for all and sundry.

4. Protection from physical diseases.

5. Protection and prevention from mental impairment.

6. To maintain the right balance between the nature’s bountiful resources and to protect them.

7. To increase the life expectancy.

8. To enjoy a stress-free high quality existence.

9. To increase the employees productivity.

10. To boost up the organisational growth at the micro level.


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11. To felicitate nations to be more competitive at a macro level.

12. To integrate and synthesise the world as one at the global level.

Control of pollution : Pollution cannot effectively be controlled by the use of a single technique. Since thecauses and the effects of different types of pollution are diverse and distinct from each other, there cannotbe any thumb- rule or golden rule which could be applied in the present context. However, in general, aprudent use of the techniques given below, in conjunction with other appropriate measures, is bound tobring in the desired results. These are:

1. Controlling at source.

2. Controlling during processes, operations and other activities.

3. Control by suitable enclosures.

4. Control by protection.

5. Control by preventions.

6. Control by absorption.

7. Adhering to regulations laid down by the following authorities:

(a) Respective state governments.

(b) Central Government.

(c) Guidelines issued by the global bodies representing individual pollution control measures.

(d) Ensuring compliance with any other law for the time being in force.

1.7 PLANT LAYOUT

Plant Layout, also known as layout of facility refers to the configuration of departments, work-centres andequipment and machinery with focus on the flow of materials or work through the production system.

Plant layout or facility layout means planning for location of all machines, equipments, utilities, workstations, customer service areas, material storage areas, tool servicing areas, tool cribs, aisles, rest rooms,lunch rooms, coffee/tea bays, offices, and computer rooms and also planning for the patterns of flow ofmaterials and people around, into and within the buildings. Layout planning involves decisions about thephysical arrangement of economic activity centres within a facility. An economic activity centre can beanything that consumes space, a person or group of people, a machine, a work station, a department, anaisle, a store room and so on. The goal or layout planning is to allow workers and equipments to operatemore effectively.

The questions to be addressed in layout planning are:

• HOW much space and capacity does each centre need?

• How should each center’s space be configured?

• What centres should the layout include?

• Where should each centre be located?


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The location of a centre has two dimensions:

• Absolute location or the particular space that the centre occupies within the facility.

• Relative location i.e., the placement of a centre relative to other centers.

The importance of layout decisions:

The need for layout planning arises both in the process of designing new plants and the redesigning existingplants or facilities.

Most common reasons for design of new layouts are:

(i) Layout is one of the key decisions that determine the long-run efficiency in operations.

(ii) Layout has many strategic implications because it establishes an organisation’s competitive prioritiesin regard to capacity, processes, flexibility and cost as well as quality of work life, customer contactand image (in case of service organisations).

(iii) An effective layout can help an organisation to achieve a strategic advantage that supportsdifferentiation, low cost, fast response or flexibility.

(iv) A well designed layout provides an economic layout that will meet the firm’s competitive requirements.

Need for redesign of layout arises because of the following reasons:

• Accidents, health hazards and low safety,

• Changes in environmental or legal requirements,

• Changes in processes, methods or equipments,

• Changes in product design/service design,

• Changes in volume of output or product-mix changes,

• Inefficient operations (high cost, bottleneck operations),

• Introduction of new products/services,

• Low employee morale.

Good Plant layout- Objectives:

• Efficient utilisation of labour reduced idle time of labour and equipments,

• Higher flexibility (to change the layout easily),

• Higher utilisation of space, equipment and people (employees),

• Improved employee morale and safe working conditions,

• Improved flow of materials, information and people (employees),

• Improved production capacity,

• Reduced congestion or reduced bottleneck centers,

• Reduced health hazards and accidents,

• Reduced material handling costs,


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• To allow ease of maintenance,

• To facilitate better coordination and face-to-face communication where needed,

• To improve productivity,

• To provide ease of supervision,

• To provide product flexibility and volume flexibility,

• To utilise available space efficiently and effectively.

Choices of Layout:

Layout choices can help greatly in communicating an organisation’s product plans and competitive priorities.Layout has many practical and strategic implications. Altering a layout can affect an organisation and howwell it meets its competitive priorities by:

• Facilitating the flow of materials and information,

• Improving communication,

• Improving employee morale,

• Increasing customer convenience and sales (in service organisations such as retail stores),

• Increasing the efficient utilisation of labour and equipment,

• Reducing hazards to employees.

The type of operations carried out in a firm determines the layout requirements.

Some of the fundamental layout choices available to managers are:

• Whether to plan the layout for the current or future needs?

• Whether to select a single-story or multistory building design?

• What type of layout to choose?

• What performance criteria to emphasise?

Factors influencing layout choices:

Primarily the layout of a plant is influenced by the relationship among materials, machinery and men.Other factors influencing layout are type of product, type of workers, the type of industry, managementpolicies etc.

Some of these factors are discussed in detailed below:

• Location: The size and type of the site selected for the plant, influences the type of buildings (singlestory or multi story) which in turn influences the layout design. Also, the location of the plantdetermines the mode of transportation from and into the plant (such as by goods trains, truck, orships) and the layout should provide facilities for mode of transport used. Also, the layout shouldprovide for storage of fuel, raw materials, future expansion needs, power generation requirementsetc.

• Machinery and Equipments: The type of product, the volume of production, type of processes andmanagement policy on technology, determines the type of machines and equipments to be installedwhich in turn influence the plant layout.


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• Managerial Policies: regarding volume of production, provision for future expansion, extent ofautomation, make-or-buy decisions, speed of delivery of goods to customers, purchasing and inventorypolicies and personnel policies influence the plant layout design.

• Materials: Plant layout includes provision for storage and handling of raw materials, supplies andcomponents used in production. The type of storage areas, racks, handling equipments such as cranes,trolleys, conveyors or pipelines etc., used - all depend on the type of materials used - such as solid,liquid, light, heavy, bulky, big, small etc.

• Product: The type of product i.e., whether the product is light or heavy, big or small, liquid or solidetc., it influences the type of layout. For example, Ship building, Aircraft assembly, Locomotiveassembly etc., requires a layout type different from that needed to produce refrigerators, cars, scooters,television sets, soaps, detergents, soft drinks etc. The manufacturing process equipments and machinesused and the processing steps largely depend on the nature of the product and hence the layoutdesign depends, very much on the product.

Type of Industry:

Figure : Type of Industry Process

I is intermittent type of industryC is continuous type of industryWhether the industry is classified under (a) Synthetic, (b) Analytical, (c) Conditioning and (d) Extractionindustries and again whether the industry has intermittent production or continuous production has arelevance to the type of layout employed.

• Workers : The gender of employees (men or women), the position of employees while working (i.e.,standing or sitting), employee facilities needed such as locker rooms, rest rooms, toilets, canteens,coffee/tea bays etc., are to be considered while designing the plant layouts.

Plant Layout- Principles:

The layout selected in conformity with layout principles should be an ideal one. These principles are:-

• Principle of Minimum Travel: Men and materials should travel the shortest distance between operationsso as to avoid waste of labour and time and minimise the cost of materials handling.

• Principle of Sequence: Machinery and operations should be arranged in a sequential order. This principleis best achieved in product layout, and efforts should be made to have it adopted in the processlayout.

• Principle of Usage: Every unit of available space should be effectively utilised. This principle shouldreceive top consideration in towns and cities where, land is costly.

Industry

Synthetic Analytical Conditioning Extractive

I C I C I C I C


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• Principle of Compactness: There should be a harmonious fusion of all the relevant factors so that thefinal layout looks well integrated and compact.

• Principle of Safety and Satisfaction: The layout should contain built in provisions for safety for theworkmen. It should also be planned on the basis of the comfort and convenience of the workmen sothat they feel satisfied.

• Principle of Flexibility: The layout should permit revisions with the least difficulty and at minimumcost.

• Principle of Minimum Investment: The layout should result in savings in fixed capital investment,not by avoiding installation of the necessary facilities but by an intensive, use of available facilities.

Types of Layout:

A layout essentially refers to the arranging and grouping of machines which are meant to produce goods.Grouping is done on different lines. The choice of a particular line depends on several factors. The methodsof grouping or the types of layout are:

(i) Process layout or functional layout or job shop layout; (ii) Product layout or line processing layout orflow-line layout; (iii) Fixed position layout or static layout; (iv) Cellular manufacturing (CM) layout orGroup Technology layout and (v) Combination layout or Hybrid layout.

Process Layout:

Also called the functional layout, layout for job lot manufacture or batch production layout, the processlayout involves a grouping together of similar machines in one department. For example, machinesperforming drilling operations are installed in the drilling department; machines performing turningoperations are grouped in the turning department; and so on. In this way, there would be an electro-plating department, a painting department, a machining departments and the like, where similar machinesor equipments are installed in the plants which follow the process layout. The process arrangement issignified by the grouping together of like machines based upon their operational characteristics. For example,centre lathes will be arranged in one department, turret lathes in a second department, and milling machinesin a third departments.

Figure of Prosses Layout

A quantity of raw material is issued to a machine which performs the first operation. This machine may besituated anywhere in the factory. For the next operation, a different machine may be required, which may


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be situated in another part of the factory. The material should be transported to the other machine for theoperation. Thus, material would move long distances and along crisscrossing paths. At one stage, thematerial may be taken to a separate building, say, for heat treatment, and then brought back for grinding.If machines in one department are engaged, the partly finished product awaiting operations may be takento the store and later reissued for production. Partly finished goods would be waiting for processing inevery department, like commuters waiting for buses in a city.

Machines in each department attend to any product that is taken to them. These machines are, therefore,called general purpose machines. Work has to be allotted to each department in such a way that no machinein any department is idle. In a batch production layout, machines are chosen to do as many different jobsas possible, i.e., the emphasis is on general purpose machines. The work which needs to be done is allocatedto the machines according to loading schedules, with the objective of ensuring that each machine is fullyloaded. The process layout carries out the functional idea of Taylor and from the historical point of view,process layout precedes product layout. This type of layout is best suited for intermittent type of production.

While grouping machines according to the process type, certain principles must be kept in mind. Theseare:

• Convenience for inspection.

• Convenience for supervision. Process layout may be advantageously used in light and heavyengineering industries, made-to-order furniture industries and the like.

• The distance between departments needs to be as short as possible with a view to avoiding long-distance movement of materials.

• Though similar machines are grouped in one department, the departments themselves should belocated in accordance with the principle of sequence of operations. For example, in a steel plant, theoperations are smelting, casting; rolling etc. These different departments may be arranged in thatorder to avoid crossovers and backtracking of materials.

Product Layout:

Also called the straight-line layout or layout for serialised manufacture, the product layout involves thearrangement of machines in one line depending upon the sequence of operations. Material is fed into thefirst machine and finished products come out of the last machine. In between, partly finished goods movefrom machine to machine, the output of one machine becoming the input for the next. In a sugar mill,sugar cane, fed at one end of the mill comes out as sugar at the other end. Similarly, in paper mill, bamboosare fed into the machine at one end and paper comes out at the other end.

In product layout, if there are more than one, line of production, there are as many, lines of machines. Theemphasis here, therefore, is on special purpose machines in contrast to general purpose machines, whichare installed in the process layout. Consequently, the investment on machines in a straight line layout ishigher than the investment on machines in a functional layout.


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Figure of Product Layout

The grouping of machines should be done, on product line, keeping in mind the following principles:

• All the machine tools or other types of equipment must be placed at the point demanded by thesequence of operations.

• All the operations, including assembly, testing and packing should be, included in the line.

• Materials may be fed where they are required for assembly but not necessarily all at one point; and

• There should be no points where one line crosses another line;

The product layout may be advantageously followed in plants manufacturing standardised products on amass scale such as chemical, paper, sugar, rubber, refineries and cement industries.

Layout in the form of Fixed Position:

As the term itself implies, the fixed position layout involves the movement of men and machines to theproduct which remains stationary. In this type of layout, the material or major component remains in afixed location, and tools, machinery and men as well as other pieces of material are brought to this location.The movement of men and machines to the product is advisable because the cost of moving them wouldbe less than the cost of moving the product which is very bulky.

Also called static layout, this type is followed in the manufacture, if bulky and heavy products, such aslocomotives, ships, boilers, air crafts and generators.

Product (Ship) Ship Assembly

Material

Machine Men

Tools

To be moved towards the material

Mixed Layout or Combined Layout:


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The application of the principles of product layout, process layout or fixed location layout in their strictmeanings is difficult to come across. A combination of the product and process layouts, with an emphasison either, is noticed in most industrial establishments. Plants are never laid out in either pure form. It ispossible to have both types of layout in an efficiently combined form if the products manufactured aresomewhat similar and not complex.

Fig. Component flow in combined layout

Layout of Service Facility:

The fundamental difference between service facility and manufacturing facility layouts is that many servicefacilities exist to bring together customers and services. Service facility layouts should provide for easyentrance to these facilities from freeways and busy thoroughfares. Large, well organized and amply lightedparking areas and well designed walkways to and from parking areas are some of the requirements ofservice facility layouts.

Because of different degree of customer contact, two types of service facility layouts emerge, viz., thosethat are almost totally designed around the customer receiving and servicing function (such as banks) andthose that are designed around the technologies, processing of physical materials and production efficiency(such as hospitals).

Other facilities with reference to Plant Layout:

A plant layout involves, besides the grouping of machinery, an arrangement for other facilities as well.Such facilities include receiving and shipping points, inspection facilities, employee facilities and storage.Not all the facilities are required in every plant. The requirements depend on the nature of the productwhich is manufactured in a particular plant.

Importance of layout:

The importance of a layout can be described as under:

• Avoidance of Bottlenecks: Bottlenecks refer to any, place in a production process where materialstend to pile up or produced at rates of speed less rapid than the previous or subsequent operations.Bottlenecks are caused by inadequate machine capacity, inadequate storage space or low speed onthe part of the operators. The results of bottlenecks are delays in production schedules, congestion,accidents and wastage of floor area. All these may be overcome with an efficient layout.


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• Avoidance of Unnecessary and Costly Changes: A planned layout avoids frequent changes which aredifficult and costly. The incorporation of flexibility elements in the layout would help in the avoidanceof revisions.

• Better Production Control: Production control is concerned with the production of a product of theright type at the right time and at reasonable cost. A good plant layout is a requisite of good productioncontrol and provides the plant control officers with a systematic basis upon which to build organisationand procedures.

• Better Supervision: A good plant layout ensures better supervision in two ways: (a) Determining thenumber of workers to be handled by a supervisor and (b) Enabling the supervisor to get a full view ofthe entire plant at one glance. A good plant layout is, therefore, the first step in good supervision.

• Economies in Handling: Nearly 30 per cent to 40 per cent of the manufacturing costs are accounted forby materials handling. Every effort should, therefore, be made to cut down en this cost. Long distancemovements should be avoided and specific handling operations must he eliminated.

• Effective Use of Available Area: Every unit of the plant area is valuable, especially in urban areas.Efforts should be therefore, be made to make use of the available area by planning the layout properly.

• Improved Employee Morale: Employee morale is achieved when workers are cheerful and confident.This state of mental condition is vital to the success of any organisation. Morale depends on betterworking conditions; better employee facilities; reduced number of accidents; and increased earnings.

• Improved Quality Control: Timely execution of orders will be meaningful when the quality of theoutput is not below expectations. To ensure quality, inspection should be conducted at different stagesof manufacture. An ideal layout provides for inspection to ensure better quality control.

• Improved Utilisation of Labour: A good plant layout is one of the factors in effective utilisation oflabour. It makes possible individual operations, the process and flow of materials handling in such away that the time of each worker is effectively spent on productive operations.

• Minimisation of Production Delays: Repeat order and new customers will be the result of promptexecution of orders. Every management should try to keep to the delivery schedules.

• Minimum Equipment Investment: Investment on equipment can be minimised by planned machinebalance and location, minimum handling distances, by the installation of general purpose machinesand by planned machine loading. A good plant layout provides all these advantages.

Good Layout - Characteristics:

• Effective coordination and integration among men, materials and machinery to maximise utilisationand output.

• Facilitates supervision and control.

• Flexibility for change of layout, expansion, changes in product design and process.

• Good working conditions - lighting, ventilation, temperature, humidity etc., are as per requirements.

• Maximum utilisation of available space.

• Proper location of storage areas.


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• Provision of safety and reduction of accidents.

• Smooth flow of production (i.e., raw materials and workers).

• Smooth movement of men, materials and machinery from place to place.

Layout planning and design:

The layout process starts with an analysis of the product to be manufactured and the expected volume ofits production. An analysis of the product includes a study of the parts to be manufactured and/or boughtand the stages at which they should be assembled to obtain the end product (i.e., finished product). Thevolume of production is estimated based on the demand and the installed capacity.

For a given product, at a desired volume of production the most appropriate process is determined. Theprocess which is selected, in turn determines the type of equipments or machinery that would be requiredto manufacture the product at ‘given volume. The type of machines or equipments could be of generalpurpose or special purpose depending on the number of products produced (i.e., a standard product or aproduct-line or product-mix) and the volume of production (i.e., batch production, mass production orcontinuous production). The type of production determines the type of layout which is most suitable tothe type of production and the process selected. The capacity planning determines the number of machinesof each type required to produce a given volume of the product or products. The modern trend is to go foradvanced manufacturing technology and automation with a trend to use capital intensive productionfacilities rather than labour-intensive. This results in increased production, better quality products, higherproductivity, shorter manufacturing cycles and reduced interruptions in production due to labour problemsetc.

Once the process and equipments needed to carryout the process are determined, the number of directoperators and indirect labour for activities such as planning, material handling, quality control , maintenance,industrial engineering, tool room, tool servicing etc., are determined. The facility planning and designmust take into consideration the space requirements for all the direct and indirect facilities including,equipments, machinery, labour, inspection areas, storage areas, utilities and services and so on. Once allthese requirements are planned, the Civil Engineers, Architects, Plant Engineering personnel all worktogether to prepare the Blue prints for the layout design of manufacturing shops and the entire plant.

Problems & Solutions

Problem: 1

A company is planning to undertake the production of medical testing equipments has to decide on thelocation of the plant. Three locations are being considered, namely, A, B and C. The fixed costs of threelocations are estimated to be Rs. 300 Lakhs, 500 Lakhs and 250 Lakhs respectively. The variable costs areRs. 3000, Rs. 2000 and Rs. 3500 per unit respectively. The average sales price of the equipment is Rs. 7000per unit. Find

(i) The range of annual production/sales volume for which each location is most suitable.

(ii) Select the best location, if the sales volume is of 18,000 units.


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Solution:

Determination of total costs of three locations.

Total cost = Fixed cost + [volume or quantity produced] × [variable cost]

= F + x.v where ‘x’ is the quantity to be produced.

a. Total cost at A = 3,00,00,000 + 3,000x ....................................................... (1)

b. Total cost at B = 5,00,00,000 + 2,000x ....................................................... (2)

c. Total cost at C = 2,00,00,000 + 3,500x ....................................................... (3)

For the various volumes of production, i.e., 5,000, 10,000, 15,000, 20,000 and 25,000 units, the total costs arecomputed at the three locations and they are plotted as shown in figure.

Table: Total costs at different volumes for three locations(Rs. in Lakhs)

Volume (Nos) 5,000 10,000 15,000 20,000 25,000

A 450 600 750 900 1,050

B 600 700 800 900 1,000

C 425 600 775 950 1,125

Decision Rules

For quantities upto 20,000 units, C is the most economical location. For quantities above 22,000, A is thepreferred location.


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Problem: 2

The present layout is shown in the figure. The manager of the department is intending to interchange thedepartments C and F in the present layout. The handling frequencies between the departments is given.All the departments are of the same size and configuration. The material handling cost per unit lengthtravel between departments is same. What will be the effect of interchange of departments C and F in thelayout?

A C E

B D F

From / To A B C D E FA – 0 90 160 50 0B – – 70 0 100 130C – – _ 20 0 0D – – – _ 180 10E – – – – 40F – – – – – –

From / To A B C D E FA 1 1 2 2 3B 2 1 3 2C 1 1 2D 2 1E 1

F –

(ii) Computation of total cost matrix (combining the inter departmental material handling frequenciesand distance matrix.

From / To A B C D E F TotalA 0 90 320 100 0 510B 140 0 300 260 700C 20 1 0 20D 360 10 370E 40 40F –

Total 1,640

If the departments are interchanged, the layout will be represented as shown below.

A C E

B D F


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The distance matrix and the cost matrix are represented as shown.

From / To A B C D E FA 1 3 2 2 1B 2 1 3 2C 1 1 2D 2 1E 1F

Total cost matrix for the modified layout.

From / To A B C D E F TotalA – 0 270 320 100 0 690B 140 0 300 260 700C 20 1 0 20D 360 10 370E 40 40F –

Total 1,820

The interchange of departments C and F increases the total material handling cost. Thus, it is not a

desirable modification.

Problem: 3

A defence contractor is evaluating its machine shops current process layout. The figure below showsthe current layout and the table shows the trip matrix for the facility. Health and safety regulationsrequire departments E and F to remain at their current positions.

E B F

A C D

Current Layout

From / To A B C D E FA 8 3 9 5B - 3C - 8 9D - 3E - 3

F -

Can layout be improved? Also evaluate using load distance (ld) score.


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Solution:

Keep the departments E and F at the current locations. Because C must be as close as possible to bothE and F, put C between them. Place A directly south of E, and B next to A. All of the heavy trafficconcerns have been accommodated. Department D is located in the remaining place. The proposedlayout is shown in figure below. The load distance (ld) scores for the existing and proposed layout areshown below. As Id score for proposed layout is less, the proposed layout indicates improvementover existing.

E C F

A B D

Proposed Layout

Dept. Pair No. of Existing plan Proposed plan

Trips Distance Load × Distance Distance Load × Distance

(1) (2) (1 × 2) (3) (1 ×3)

A–B 8 2 16 1 8

A–C 3 1 3 2 6

A–E 9 1 9 1 9

A–F 5 3 15 3 15

B–D 3 2 6 1 3

C– E 8 2 16 1 8

C–F 9 2 18 1 9

D–F 3 1 3 1 3

E–F 3 2 6 2 6

Total 92 67

Problem: 4Location A would result in annual fixed cost of Rs. 3,00,000 variable costs of Rs. 63 per unit andrevenue Rs. 68 per unit. Annual fixed cost at Location B are Rs. 8,00,000 variable costs are Rs. 32 perunit and revenues are Rs. 68 per unit. Sales volume is estimated to be 25000 units/year, which locationis attractive?

Solution:

Location A: BEP (units) =60,000 and Location B: BEP (units) =22,222.

At the expected demand of 25000 units, profits (loss) for the alternatives are:


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Alternatives

Location B is most attractive, even though annual fixed costs are much higher than A.

Problem: 5A manufacturer is considering four locations for a new plant. It has attempted to study all costs at thevarious locations and find that the costs of the following items vary from one location to another. Thefirm will finance the new plant from deposits bearing 10 percent interest.

A B C DLabour (Rs. per unit) 0.75 1.10 0.80 0.90

Plant (Rs. crores) 0.46 0.39 0.40 0.48

Materials & equipment * (Rs. per unit) 0.43 0.60 0.40 0.55

Electricity (per year) (Rs.) 30.00 26.00 30.00 28.00

Water (per year) (Rs.) 7.00 6.00 7.00 7.00

Transportation (per unit) (Rs.) 0.02 0.10 0.10 0.05

Taxes (per year) (Rs.) 33.00 28.00 63.00 35.00

* This cost includes a projected depreciation, but no interest.

Determine the most suitable location (economically) for output volumes in the range of 50,000 to1,30,000 units per year.

Costs A B C D

Fixed Cost (per year) :10% of investment 4,60,000 3,90,000 4,00,000 4,80,000Electricity 30,000 26,000 30,000 28,000Water 7,000 6,000 7,000 7,000Taxes 33,000 28,000 63,000 35,000

Total Fixed Cost 5,30,000 4,50,000 5,00,000 5,50,000

Variable Cost :Labour 0.75 1.10 0.80 0.90Material & equipment 0.43 0.60 0.40 0.55Transportation 0.02 0.10 0.10 0.05

Total Variable Cost (per unit) 1.20 1.80 1.30 1.50

Total Cost: 5,30,000 4,50,000 5,00,000 5,50,000

+ 1.2/unit + 1.8/unit + 1.3/unit 1.5/unit

A BRevenue 1,700,000 1,700,000Costs:Variables 1,575,000 800,000Fixed 300,000 800,000

1,875,000 1,600,000Profit/(Loss) (175,000) 100,000


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The point for a plant plans location break-even analysis chart are as follows:

At zero unit of output, use fixed cost values. At 100000 units of output:

A = 6,50,000; B = 6,30,000; C = 6,30,000; D = 7,000,000.

Plot the graph of these linear functions.

Solution:

From the graph it is clear that, for minimum cost, use site B for a volume of 50,000 to 1,00,000 units,use site C for a volume of 1,00,000 to 1,30,000 units.

Problem: 6

Mr. X, the factory Manager of S.K.Industries, is considering an interchange of departments 3 and 6 in thepresent layout. The present layout and the interdepartmental materials handling frequencies are furnishedbelow [All the departments are of the same size and configuration in the following matrix, respectively.]

1 3 5

2 4 6

Present LayoutWeekly frequencies of interdepartmental Material handling

From / To 1 2 3 4 5 61 – 0 90 160 50 62 – – 70 0 100 1303 – – _ 20 0 04 – – – _ 180 105 – – – – 40

The per unit length inter departmental cost of materials of materials handling are equal. What is the effectof the interchange of the departments 3 and 6 in the layout?

Solution:

The distance matrix for the present layout can be given as follows (Considering only the departmentsthat share a border as adjacent departments).


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Distance matrix (for initial layout)

From / To 1 2 3 4 5 61 1 1 2 2 32 – 2 1 3 23 – – 1 1 24 – – – 2 15 – – – 1

6 – – – – –

The total cost matrix can be easily calculated combining the inter-departmental materials handlingfrequencies and the distance matrices.

Total cost matrix (for the initial layout)

From / To 1 2 3 4 5 6

1 0 90 320 100 0 5102 140 0 300 260 7003 20 0 0 204 360 10 3705 40 406

Total 1,640

If the department 3 and 6 are interchanged, the layout would be as follows:

1 6 5

2 4 3

From / To 1 2 3 4 5 61 – 1 3 2 2 12 – – 2 1 3 23 – – _ 1 1 24 – – – _ 2 15 – – – – 1

From / To 1 2 3 4 5 6 TotalA 0 270 320 100 0 690B 140 0 300 260 700C 20 1 0 20D 360 10 370E 40 40F –

Total 1,820


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By interchanging the departments 3 and 6 the total cost has been increased. So it is not advisable.

Problem: 6

A company planning, to manufacture a household cooking range, has to decide on thelocation of the plant. Three locations are being considered viz., Patna, Ranchi, and Dhanbad. The fixedcosts of the three locations are estimated to be Rs. 30 lakh, 50 lakh, and 25 lakh per annum respectively. Thevariable costs are Rs.300, Rs.200 and Rs.350 per unit respectively.

The expected sales price of the cooking range is Rs.700 per unit find out

(a) the range of annual production/sales volume for which each location is the most suitable and

(b) Which one of the three locations is the best location at a production/sales volume of 18,000 units?

Solution:The total cost of the three locations are:

At Total cost = Fixed cost + Variable cost for a volume “X”

Patna => Total cost = 30,00,000 + 300 × XRanchi => Total cost = 50,00,000 + 200 × XDhanbad => Total cost = 25,00,000 +350 ×XWe can compute and plot the total costs per annum at the three different locations for the variouscases of production volume of 5000, 10000, 15000, 20000, 25000 units.

a) Patna Volume (Units) 5,000 10,000 15,000 20,000 25,000 Total =30,00,000+ 30,00,000+ 30,00,000+ 30,00,000+ 30,00,000+ Cost (Rs.) 300(5,000) 300(10,000) 300(15,000) 300(20,000) 300(25,000) =Rs.45 lakhs =Rs.60 lakhs =Rs.75 lakhs =Rs.90 lakhs =Rs.105lakhs b) Ranchi Volume (Units) 5,000 10,000 15,000 20,000 25,000 Total = 50,00,000+ 50,00,000+ 50,00,000+ 50,00,000+ 50,00,000+ Cost (Rs.) 200(5,000) 200(10,000) 200(15,000) 200(20,000) 200(25,000) =Rs.60

lakhs =Rs.70

lakhs =Rs.80 lakhs

=Rs.90 lakhs

=Rs. 100 lakhs

c) Dhanbad Volume (Units) 5000 10000 15000 20000 25000 Total = 25,00,000+ 25,00,000+ 25,00,000+ 25,00,000+ 25,00,000+ Cost (Rs.) 350(5,000) 350(10,000) 350(15,000) 350(20,000) 350(25,000) =Rs.42.5

1akhs=Rs.601akhs

=Rs.77.5 1akhs

=Rs.95 1akhs

=Rs.112.5 1akhs

If the volume distribution be as follows: up to 10,000

unit between 10,000 units

to 20,000 units above 20,000 units

Favourable Location

Dhanbad Patna Ranchi


46

Cost Matrix From To A B C

A 1 1 B 1 1 C 1 1

Flow Matrix Distance Matrix To A B C To A B C

From From A 1 2 A 1 2 B 1 3 B 1 1 C 3 3 C 2 1

For a volume 18000 units favourable location is Patna which can be substantiated by the followings:

Patna => 30,00,000 +300 × 18000 = Rs. 84 lakhsRanchi => 50,00,000+200 × 18000 = Rs. 86 lakhsDhanbad => 25,00,000 +350 × 18000 = Rs. 88 lakhs

Problem: 7Find an improved layout for the initial layout given in figure by using CRAFT pair wise exchangetechnique. The interdepartmental flows are also furnished along with the interdepartmental cost matrix.

From To A B C Total A 1 4 5 B 1 3 4 C 6 3 9

Total Cost 18

Solution:

For the initial layout let us find the total cost matrix (which is the product of the cost flow and distancematrices given).

Total Cost Matrix (Initial)

CRAFT considers exchanges between a pair of departments which have

i) either a common border, or ii) the same area.

Since in the given problem, all the three departments A, B and C have the same area they can beinterchanged with each other in pairs. The interchanges that are possible are between, (i) A and B, (ii)

A B C

11 1

Initial Layout

1


47

Distance Matrix To

From A B C A 1 1 B 1 2 C 1 2 Therefore, the total cost matrix will be

Cost Matrix To

From A B C Total A 1 2 3 B 1 6 7 C 3 6 9 Total Cost = 19

The total cost is now more than that for the initial layout and so this exchange is not desirable. Wemay next try the exchange between A and C. However, this exchange results in only a mirror imageof the initial layout. As such, that is not giving to change the total cost, and layout. Therefore we trythe next possible exchange, viz., between B and C (Figure).

A C B

This matrix is an improvement over that for the initial layout and is thus accepted. The improvedlayout flow is shown in figure.

To From A B C Total A 2 2 4 B 2 3 5 C 3 3 6 Total 5 5 5 15

A and C, and (iii) B and C.

Let us interchange A and B (Figure).

New LayoutNew Distance MatrixThe total cost matrix is given belowTotal Cost Matrix

A B C

11 1

Initial Layout

1


48

Final Layout

Material Handling System:

Material handling is an integral part of manufacturing process. In order to manufacture any product, it isessential to move the material from one place to that of another. Materials must be turned, moved, andpositioned on the respective machines to produce the required output. Materials must also be inevitablymoved from one machine to another. Though operators can be employed to move the materials, the commonpractice is to move the materials from one processing area to another and from one department to another.To start with, material must-be moved, prior to the production process, from the storage room to the firstprocessing operation. There may also be inter-departmental transfer of materials or finned goods from thefinal conversion operation performed to final inspection. Finally, they go to the finished goods room andthen shipped to customers.

In a general sense, materials handling includes all movement of materials in a manufacturing situation.Materials handling is defined as the art and science involving the moving, packing, and storing of substancesin any form.

The primary objective of materials handling is to improve production performance by speeding up materialflow. Lower unit cost of production less work in progress, reduced labour costs of handling materials,better utilization of shop floor and warehouse, fewer breakages, reduced fatigue of operatives and othershop floor personnel, better facilities and reduced accidents, improved product quality, better customerservice, and general safety are the other objectives of the materials handling system. The twin objectivestherefore can be:

• Improvement of production performance and reduction of costs.

• Improvement of conditions for shop floor personnel.

These are discussed as under:

• Better control of the flow of goods: Effective materials handling system results in better control of theflow of goods.

• Higher productivity at lower manufacturing cost: The material handling system in any manufacturingorganization should be primarily designed to improve productivity and avoid inordinate delays infurnishing the required materials at the manufacturing place. The fastest most efficient and economicalmovement of materials result in higher productivity at lower manufacturing cost.

• Improved working conditions and greater safety in the movement of materials: The manufacturingorganizations are required to follow the safe handling practices. Safe handling of materials limits thecases of industrial accidents and employees feel safe and secure to work in manufacturing units.

• Increased storage capacity: Scientific movement and storage of materials result in the effectiveutilization of available storage space.

• Lower the unit materials handling costs: It is quite obvious that the overall materials handling costswill be reduced if the unit costs are reduced.

A C B


49

• Provide for fewer rejects: Careful handling of the product will contribute to a better quality level ofthe goods, produced. Ineffective handling of materials results in breakage, and damage, to the goodsproduced. Production of defective pieces acts as a major cost to manufactures.

• Reduce the manufacture cycle time: Through effective handling of materials, the total time required tomake a product move from the receipt of raw materials to the finished goods. Movement of materialscan be speeded up or may be handled over shorter distances. Substantial reduction of the manufacturingcycle time will eventually reduce the inventory costs and the other production costs incident thereto.

Principles of Material Handling

Materials handling is a service function; it is not an end in itself. The principles of materials handling canbe grouped under three broad headings:

• Principles relating to the elimination of wasteful methods,

• Principles relating to the laying out the plant,

• Principles relating to the selection and application of materials handling equipment.

Wasteful methods can be eliminated by following the undermentioned principles.

• Avoiding the unnecessary transfer of materials from floor to workplace or from container to container,

• Eliminating unnecessary mixing and subsequent storing,

• Increasing the speed of handling the materials,

• Utilizing gravity as a moving force, wherever practicable,

• Introduction of automaticity into the materials handling system,

• Reducing to a minimum the number of handlings of materials,

• Using mechanical aids to eliminate the use of hand labour in the movement of materials.

Principles of Material Handling

Related to Planning

Related to Equipment

Related to Operation

• Planning Principle • Systems Principle • Material Flow Principles • Simplification Principle • Gravity Principle • Space Utilisation Principle • Safety Principle

• Mechanisation/ Automation Principle • Equipment Selection Principle • Flexibility Principle • Dead Weight Principle • Motion Principle • Idle Time Principle • Maintenance Principle • Obsolesence Principle

• Control Principle • Capacity Principle • Performance Principle

Fig. Principles of Material Handling


50

Selection of Materials handling equipment

The following principles should be borne in mind; while, selecting and applying various materials handlingequipment:

• As far as possible, management should avoid; using the complicated mechanisms and controls; it isbetter, to use the simplest possible equipment to handle the materials.

• Before considering the purchase of new equipment; the activities must be planned and the equipmentneeds must be analysed.

• Before purchasing the equipment, comparative costs of various equipments must be determined andanalysed.

• Different equipment must be used for different jobs.

• Equipment selected for handling materials must be flexible in its application.

• It is necessary that equipment must be managed and maintained properly.

• It is necessary to ensure that obsolete methods and equipment are replaced periodically.

• Management should ensure that the new handling equipment must be put to effective use.

• Management should try to standardize equipment, if possible.

• Selection of equipment should be based on the principle that the equipment should minimize theratio of mobile equipment dead weights to payloads.

• The building must big enough to keep the materials handling equipment.

Various types of materials handling equipment used in manufacturing organisations:

Materials handling requires some equipment especially designed to meet the requirements of handlingdifferent materials. The general types of materials handling equipment include the conveyors, cranes,elevator and hoists; positioning; weighing and control equipment; industrial vehicles motor vehicles; railroadcars; marine carriers; aircraft; containers and supports. The main types of equipment can be convenientlyclassified under the following heads;

• Aerial ropeways and cable-ways

• Conveyors: These include

(i) Chutes

(ii) Roller Conveyors

(iii) Belt conveyors

(iv) Drag lines

(v) Bar or slot

(vi) Tow

(vii) Screw

(viii) Pneumatic.

(ix) Bracket or tray elevators

(x) Car and chair conveyors


51

(xi) Monorails

The conveyors are expensive to install, but cheap to run. Further, they are reliable in operation. Most of theconveyors are inflexible and restrict plant layout changes. They have one positive advantage - they requirelittle maintenance.

• Hoists and lifts

• Cranes: Actually, it is very difficult to precisely define cranes as they shade in one direction into hoistsand in the other direction into conveyors. Several possibilities could be:

(i) Portable or mobile

(ii) Fixed track travelling

(iii) Fixed

(iv) Other sub-types including the job, gantry, bridge, and derrick.

• Trucks and tractors: These include

(i) Straddle

(ii) Pedestrian controlled

(iii) Hand operated

(iv) Industrial tractors

• Pallet handling trucks and pallets: A pallet is a portable platform on which goods are placed to form a‘unit’ load for handling and stacking. A variety of trucks has been designed for handling pallets,including the following:

(i) Forklift trucks: A generic term covering all types of trucks capable of using forks to lift pallets.

(ii) Reach trucks: These are used where the forks are telescopic to facilitate stacking.

(iii) Stackers: These are designed in such a way that they can be operated from batteries or mainsin a fixed position.

(iv) Hand pallet trucks

(v) Stillage trucks: A stillage is a simple form of pallet, usually disposable and therefore cheaplyconstructed. Stillage trucks are usually hand-operated trucks.

The devices that can be used with forklift trucks include:

(i) Side shifter attachments for correct lateral positioning,

(ii) Special loading and off-loading attachments,

(iii) Clamping arms and grabs,

(iv) Tilting, rocking and rollover devices.

• Earth moving equipment: These are of little concern in material handling.

• Various miscellaneous equipment: These include:

(i) Lorry floor conveyors,


52

(ii) Van loader,

(iii) Lorry loader,

(iv) Lifting tail gates,

(v) Vibrating screens,

(vi) Hydraulic bridges.

• Automatic transfer equipment: The quest for automation has resulted in rapid development ofautomatic transfer equipment in the recent past.

Factors affecting the selection of materials handling equipment:

The primary objective of employing the materials handling equipment is to arrive at the lowest cost perunit of material handled. This entails attaining a happy balance between the production problem, thecapabilities of the equipment available, and the people involved in using the equipment. The factorsconcerning the production problem include the volume of production, the types and classes of materialsneeded, the layout of the plant and building premises. The factors to be taken into account while employingthe equipment for handling the materials includes:

• Adaptability: The load-carrying and movement characteristics of the equipment selected should fitthe materials handling problem.

• Cost: The most important factor in the selection of materials handling equipment is the cost of theequipment. The cost must not be prohibitive; rather it should be reasonable. A manager should considerseveral factors here; the initial purchase price, direct and indirect labour required in operating theequipment, installation costs, maintenance costs, power requirements, insurance requirements, spacerequirements, depreciation charges, salvage value, time value of money invested, and any other costs,that are to be incurred in future of course, it is expected that the anticipated annual savings frominstalling the materials handling equipment must exceed the total annual costs of investment.

• Ease of maintenance: The maintenance of equipment selected and purchased at reasonable cost isanother important factor.

• Environment: Equipment selected must conform to the environmental requirements of an organization.

• Flexibility: It is better if the equipment selected is flexible enough to handle more than one type ofmaterial.

• Power: Enough power should be available to do the job.

• Space requirements: While selecting the equipment space required to install shall have to be takeninto account.

• Speed: Within the limits of the production process of plant safely, materials must be moved at areasonable speed.

• Supervision: Depending on the degree of automaticity, supervision must be exercised on the machinesinstalled to handle the materials.

• The load capacity: The load capacity of the selected equipment must be adequate enough to performthe job effectively. However, the equipment should not be too heavy and result in excessive operatingcosts.

ProductionPlanning

andProductivityManagement

STUDY NOTE - 2

54

Production Planning and Productivity Management

2.0 Time Study, Work Study, Method Study & Job Evaluation

Time Study:

Time study is defined to be a searching scientific analysis of the methods and equipment used or plannedin doing a piece of work, development in practical detail of the best manner of doing it and determinationof the time required.

Operation analysis is the study of the entire process to determine whether operations can be eliminated,combined or the sequence changed. Operation analysis aims to determine the one best way and can beapplied to method, materials, tools equipment layout, working conditions and human requirements ofeach operation.

Job standardisation consists in determining the one best way of performing a job under the means atcommand of recording the exact method along with the time for each element of operation and establishingmeans to maintain the standard conditions.

Another term connected with time and motion study is the job analysis. Job analysis is the determinationof essential factors in a specific kind of work and of the qualifications of a worker necessary for itsperformance.

Time study aims at determining the best manner of doing a job and timing the performance of the jobwhen done in the best manner.

In motion study the work is divided into fundamental motions and in time study work is divided intoelements of operations. In both cases attempts are made to remove useless motions and improve combinationand sequences of motions and operations. In motion study the best way of doing a work is determined bymotion analysis and operators are trained to follow the method so determined but in time study the bestmethod is determined by analysis of the methods and equipment, used and motions only roughly consideredand that too indirectly. In time study, setting of production standards, standards for cost purposes andwage incentives are emphasised. The measurement of human effort is a difficult job which can only besolved by using scientific method and industrial experience combined with knowledge of psychology.

The use of scientific method involves experiment measurement and elimination of variables connectedwith a job.

The variables connected with a job are the method of manufacture, tools and equipments, material, workingconditions, worker concerned and time required to perform the job. In order to measure the last variabletime, the other variables must be eliminated by standardising. In going to proceed for time study, it is firstnecessary to standardise the method and conditions of work and to define what an average worker is.Time study has two sides, mechanical and human.

Before commencing the time study, the time study man should ensure and ascertain the following:

• That motion studies have been carried out so that planning of work, work places and appliances aresatisfactory.

• That the operations can be performed in the correct; sequence without interruption.

• That the human effort involved is minimum.

• That the worker in question has average skill and diligence necessary for the work.


55

Work Study:

It is a general term for the techniques: methods study and work measurement which are used in theexamination; of human work in all its contexts and systematically investigate all factors leading toimprovement of efficiency.

Work study aims at finding the best and most efficient way of using the available resources—men, materials,money and machinery. Once the method study has developed an improved procedure for doing a workthe work measurement or time study will study the time to complete a job.

Method Study:

It is the systematic investigation of the existing method of doing a job in order to develop and install aneasy, rapid, efficient, effective and less fatiguing procedure for doing the same job and at minimum cost.This is achieved by eliminating unnecessary motions involved in a certain operation or by changing thesequence of operation or the process itself.

Methods study can be made by the help of both motion study and time study.

The methods study programme must include the following features:–

(a) Uniform application,

(b) Established standard practice,

(c) Continuous review,

(d) Credit distribution.

A new and improved method developed in one department should be spread out to the entire plantpreferably with further improvements.

A new method must not be forgotten between orders as it happens sometimes in batch production. Methodsdepartment should always aim at improved and better ways of doing jobs.

For successful control of methods study, the enthusiastic cooperation of every employee is required. Togain employee cooperation, distribution of credit is essential. It has been correctly said that a good methodsdepartment rarely takes credit for an original idea. Its success lies in getting new ways and methods adoptedpromptly, universally, continuously and cooperatively towards the improvement of productivity.

Job Evaluation:

Job evaluation is the ranking grading, and weighing of essential work characteristics of all jobs in order tofind out or rate the worth of jobs. It is a systematic approach to ascertain the labour worth of each job andis a very important concern of all employers.

Job evaluation aims at fairness and consistency so far as all wages and salaries are concerned within anorganisation and when systematic and impartial, it stimulates, confidence of the employees. There arethree steps for evaluations of all jobs :–

(i) Preparation of preliminary description of each existing job.

(ii) Analysing each job to arrive at final job descriptions and specifications.

(iii) Analysing each job according to its approved description in order to determine its worth or value.

56


Job Description and Specifications: The understanding of the job content or job description is the primaryrequirement.

Job specifications are derived from the job descriptions which have already been approved. The specificationhelp determining the qualification required of the individual desired for the position. This in turn guidesthe personnel department in the selection of employees and also guides shop executives in the placementof workmen.

Systems of Valuation: There are several systems of job evaluation.

The fundamental criteria in valuation of a job into account are to make a specific list of factors which affectjob values. The many factors are:

(i) Qualifications required of the worker,

(ii) Job difficulties,

(iii) Job responsibilities,

(iv) Working conditions.

All these factors are to be analysed in detail in order to complete the job description. The list of factors, themanner in which they are apprised and the method of finding out relative worth and money valuesdistinguishes the various systems of valuation.

The systems of valuations which are commonly adopted are given below:

1. The ranking or grading method,

2. The factor comparison method,

3. Point rating method.

Ranking or Grading Method: Under this system the titles of all jobs are written on cards and the grading isdone by several competent judges. The hourly rates to be paid for different jobs are suggested by thejudges without any consideration to the existing wage. The ranks or grades assigned to each job by all thejudges are averaged and this average is considered the “score” for that job. Hourly rates are then fixed forjobs according to their ranking.

Factor Comparison Method: The factor comparison method analyses the job into much greater detail thanthe grading method. It ranks each job with respect to each factor that characterise the job and the factorsare taken one at a time.

All jobs are compared and ranked first with respect to mental requirements, then skill, then physicalrequirements and after that responsibility and lastly working conditions. The total worth of the job isobtained by adding together money values which are assigned separately to the various levels of rank ineach factor. Factor comparison method is more accurate than the simple ranking systems, since the separatefactors are analysed comparatively. This method is flexible.

Point Rating Method: There are three methods of analytical evaluation of a job. They are:

1. Straight point method.

2. Weighted point method.

3. Valuation of jobs directly in money method, not specifying any maximum weight.


57

Straight Point Method: This method assigns equal weights for each characteristic. When evaluating a jobunder this system, it is assumed that all the characteristics have ranges of values between same maximumand minimum points.

Weighted Point Method: In this method different points are assigned to the different characteristics ofdoing jobs.

Direct to Money Methods: After selecting the job characteristics, ten key jobs whose rates are believed tobe correct, are taken and the present wage rates of these jobs are distributed to the job characteristics byeach analyst. The jobs are then ranked by the analysts for each characteristic in order of the degree to whichthat characteristic is present. This serves as a check to show up any errors made in the original distributionof the wages rate to the various characteristics.

Problems and Solutions

Problem: 1

Continuous stopwatch study observations for a job are given. Compute the standard time for the job, if thetotal allowances are 15%.

Ele. Description Cycle time (min) P.R.

No. 1 2 3 4 5 6 7 8 9 10

A Loosen vice 0.09 0.49 0.89 1.31 1.70 2.09 2.50 2.88 3.29 3.71 90

B Set bar length 0.16 0.56 1.38 1.38 1.76 2.16 2.57 2.95 3.36 3.78 110

C Switch m/c 0.28 0.67 1.49 1.49 1.88 2.28 2.68 3.07 3.40 3.90 120

D Unlock arm & 0.41 0.80 1.61 1.61 2.00 2.41 2.80 3.20 3.62 4.03 100set saw

Solution:The individual element cycle timing is computed from the cumulative cycle times as shown in tablebelow:

Ele. Cycle time (min) Avg. Normal

No. 1 2 3 4 5 6 7 8 9 10 time time

A 0.09 0.08 0.09 0.10 0.09 0.09 0.09 0.08 0.09 0.09 0.089 0.080

B 0.07 0.07 0.06 0.07 0.06 0.07 0.07 0.07 0.07 0.07 0.068 0.075

C 0.12 0.11 0.12 0.11 0.12 0.12 0.11 0.12 0.13 0.12 0.118 0.142

D 0.13 0.13 0.14 0.12 0.12 0.13 0.12 0.13 0.13 0.13 0.128 0.128

Total 0.425

Standard time = 0.425

1 0.15− = 0.500 minutes.

58


Problem: 2

The work - study engineer carries out the work sampling study. The following observations were made fora machine shop.

Total number of observations 7000

No. Working activities 1200

Ratio between manual to machine elements 2:1

Average rating factor 120%

Total number of jobs produced during study 800 units

Rest and personal allowances 17%

Compute the standard time for the job.

Solution:

(i) Overall time per unit (To) =Duration of study

Number of jobs produced during study =120 60

800×

= 9 min.

(ii) Effective time per piece (Te) = To × Productive observations

Total observations = 9 ×

58007000

= 7.46 min.

The effective time is to be segregated into manual time and machine element time.

Machine controlled time per piece (Tm) = 7.46 × 1/3 = 2.49 min

Hand controlled time per piece (Th) = 7.46 × 2/3 = 4.97 min

Normal time per piece = Tm + Th × performance rating = 2.49 + 4.97 × 1.2 = 8.46 min.

Standard time per piece = 8.46 (1 + 0.17) = 9.9 minutes.

Problem: 3

The time study of a machinery operation recorded cycle times of 8.0, 7.0, 8.0 and 9.0 minutes. The analystrated the observed worker as 90%. The firm uses a 0.15 allowance fraction. Compute the standard time.

Solution:

Average cycle time =8.0 7.0 8.0 9.0

4+ + +

= 8.0 minutes.

Normal time = 8.0 × 0.9 = 7.2 minutes.

Standard time = 7.2

(1-0.15) = 8.47 minutes.

The standard time for this machinery operation would be set at 8.47 minutes, which is greater thanthe average cycle time observed. The average cycle time was adjusted for the rating factor (90%) andthe allowance fraction (0.15).


59

Problem: 4

An analyst wants to obtain a cycle time estimate that is within ± 5% of the true value. A preliminary run of20 cycles took 40 minutes to complete and had a calculated standard deviation of 0.3 minutes. What is thecoefficient of variation to be used for computing the sample size for the forthcoming time study?

Solution:

Standard deviation of sample(s) = 0.3 min/cycle.

Mean of sample = x = 40 min 20 cycle = 2 min./cycle;

V = sx

= 0.32

= 0.15

Problem: 5

A job has been time standard for 20 observations. The mean actual time was 5.83 minutes and the standarddeviation of the time is estimated to be 2.04 minutes. How many total observations should be taken for95% confidence that the mean actual time has been determined within 10%?

Solution:

n = 2Zs

Ax⎛ ⎞⎜ ⎟⎝ ⎠

= ( )( )

21.96 2.040.10 5.83

⎡ ⎤⎢ ⎥⎣ ⎦

= 47

Therefore, a total of 47 observations should be made. Since 20 observations have already been made,only 27 more are necessary.

Problem: 6

An analyst has observed a job long enough to become familiar with it and has divided it into five elements.The element times for the first four cycles and a performance rating for each element are given in thefollowing table,

Element Cycle Cycle Cycle Cycle Performance1 2 3 4 Rating (%)

1 1.246 1.328 1.298 1.306 90

2 0.972 0.895 0.798 0.919 100

3 0.914 1.875 1.964 1.972 100

4 2.121 2.198 2.146 2.421 110

5 1.253 1.175 1.413 2.218 100

(a) Do any of the times look like outliners, i.e. probable errors in reading or recording data thatshould not be included in the analysis?

(b) Compute an estimated normal time for the job based on the data available at this stage of the study.

(c) On the basis of the data available, what sample size should be taken to estimate the time forelement 2 within 5% of the true mean time with 95% confidence?

60


Solution:

(a) The times for element 3 in cycle 1 and for element 5 in cycle 4 are suspect and should bedisregarded.

(b) The following estimates are made on the basis of the remaining times

Element Mean actual time Performance Rating (%) Normal time

1 1.295 90 1.116

2 0.896 100 0.896

3 1.937 100 1.937

4 2.222 110 2.444

5 1.28 100 1.28

Normal time for total job = 7.723

(c) For element 2:

x = 0.896

S = ( )2

2

1

xx

nn

−′

′ −

∑∑ =

( )23.227174 – 3.584 43

= 0.0728

n = 2Zs

Ax⎛ ⎞⎜ ⎟⎝ ⎠

= ( )( )

21.96 0.07280.05 0.896

⎡ ⎤⎢ ⎥⎣ ⎦

= 10.14

The analyst probably would want to use more than 10 observation, so that workers would have moreconfidence in the standard. A Company might make it a general practice to use at least say 15 or moreobservations.

Problem: 7

Stopwatch time study figure for a job which is continuous in nature are given below. Calculate the StandardTime for the job assuming that the sample size is adequate, and total allowances are 15 percent.

Ele. Description Cycle time (min) P.R.

No. per cycle 1 2 3 4 5 6 7 8 9 10

1 A 0.10 0.50 0.90 1.32 1.71 2.10 2.51 2.89 3.30 3.72 90

2 B 0.17 0.57 0.96 1.39 1.77 2.17 2.58 2.96 3.37 3.79 110

3 C 0.29 0.68 1.08 1.50 1.89 2.29 2.69 3.08 3.41 3.91 120

4 B 0.15 0.81 1.22 1.62 2.01 2.42 2.81 3.21 3.63 4.04 100

Solution:

From the continuous study figure the individual time figures are derived.


61

Cycle time (min) NormalEle. Description 1 2 3 4 5 6 7 8 9 10 Arithmetic time ofNo. per cycle Average element

(min)

1 A 0.10 0.08 0.09 0.10 0.09 0.09 0.09 0.08 0.09 0.09 0.090 0.081

2 B 0.07 0.07 0.06 0.07 0.06 0.07 0.07 0.07 0.07 0.07 0.068 0.075

3 C 0.12 0.11 0.12 0.11 0.12 0.12 0.11 0.12 0.13 0.12 0.118 0.142

4 D 0.13 0.13 0.14 0.12 0.12 0.13 0.12 0.13 0.13 0.13 0.128 0.128

Total time = 0.426 min

Standard time = 0.426 ÷ (1 - 0.15) = 0.501 min

2.1 Production Planning and Control Introduction

Production planning control can be viewed as the nervous system of a production operation. The primaryconcern of production planning and control is the delivery of products to customers or to inventory stocksaccording to some predetermined schedule. All the activities in the manufacturing or production cyclemust be planned, coordinated, organised, and controlled to achieve this objective. From a long-term pointof view (usually from seven to ten years or more) production planning largely deals with plant constructionand location and with product-line, design and development. Short-range planning (from several monthsto a year) focuses on such areas as inventory goals and wage budgets. In plans projected over a two-to-fiveyear period, capital-equipment budgeting and plant capacity and layout are the major concern. Productionplanning and control normally reflects the short range activities and focuses on the issues and problemsthat arise in the planned utilisation of the labour force, materials, and physical facilities that are requiredfor manufacturing the products in accordance with the primary objectives of the firm.

Production systems are usually designed to produce a variety of products and are, therefore, complex. Insuch complex systems, anything can happen and usually it is so. Therefore, it is vital to exercise some kindof control over the production activities. Control is possible only when everything is planned. Productionplanning and control is thus a very important aspect of production management.

Objectives of production planning and control

The ultimate objective of production planning and control is to contribute to the profits of the enterprise.This is accomplished by keeping the customers satisfied through the meeting of delivery schedules. Further,the specific objectives of production planning and control are to establish the routes and schedules forwork that will ensure the optimum utilisation of raw materials, labourers, and machines to provide themeans for ensuring the operation of the plant in accordance with these plans. Production planning andcontrol is essentially concerned with the control of work-in-process. To control work-in-process effectivelyit becomes necessary to control not only the flow of material but also the utilisation of people and machines.

Production planning and control fulfils these objectives by focusing on the following points:

� Analysing the orders to determine the raw materials and parts that will be required for their completion,

� Answering questions from customers and salesmen concerning the status of their orders,

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� Assisting the costing department in making cost estimates of orders,

� Assisting the human resource departments in the manpower planning and assignment of men toparticular jobs,

� Controlling the stock of finished parts and products,

� Determining the necessary tools required for manufacturing,

� Direction and control of the movement of materials through production process,

� Initiating changes in orders as requested by customers while orders are in process,

� Issuing requisitions for the purchase of necessary materials,

� Issuing requisitions for the purchase or manufacture of necessary tools and parts,

� Keeping the up-to-date records scheduled and in process,

� Maintaining stocks of materials and parts,

� Notifying sales and accounting of the acceptance of orders in terms of production feasibility,

� Preparing the route sheets and schedules showing the sequence of operation required to produceparticular products,

� Production of work orders to initiate production activities,

� Receiving and evaluating reports of progress on particular orders and initiating corrective action, ifnecessary,

� Receiving orders from customers,

� Revising plans when production activities cannot conform to original plans and when revisions inscheduled production are necessary because of rush orders.

Production control involves the following functions:

� Planning the production operations in detail,

� Routing, i.e., laying down the path for the work to follow and the order in which the various operationswill be carried out,

� Scheduling, i.e., establishing the quantity of work to be done, and fixing the time table for performingthe operations,

� Dispatching, i.e., issuing the necessary orders, and taking necessary steps to ensure that the timetargets set in the schedules are effectively achieved,

� Follow-up, taking necessary steps to check up whether work proceeds according to predeterminedplans and how far there are variances from the standards set earlier,

� Inspection, i.e., conducting occasional check-ups of the products manufactured or assembled to ensurehigh quality of the production.


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Figure : Techniques of Production Control

Basic types of production control:

Production control can be of six types:

• Block control

This type of control is most prominent in textiles and book and magazine printing. In these industriesit is necessary to keep things separated and this is the fundamental reason why industries resort toblock control.

• Flow control

This type of control is commonly applied in industries like chemicals, petroleum, glass, and someareas of food manufacturing and processing. Once the production system is thoroughly designed, theproduction planning and control department controls the rate of flow of work into the system andchecks it as it comes out of the system. But, under this method, routing and scheduling are done whenthe plant is laid out. That is to say, the production line which is established is well balanced andsequenced before production operations begin; this type of control is more prevalent in continuousproduction systems.

• Load control

Load control is typically found wherever a particular bottleneck machine exists in the process ofmanufacturing.

• Order control

The most, common type of production control is called order control. This type of control is commonlyemployed in companies with intermittent production systems, the so-called job-lot shops. Under thismethod, orders come into the shop for different quantities for different products. Therefore, productionplanning and control must be based, on the individual orders.

• Special project control

Special production control is necessary in certain projects like the construction of bridges, officebuildings, schools, colleges, universities, hospitals and any other construction industries. Under thistype of control, instead of having sets of elaborate forms for tooling and scheduling, a man or a groupof men keeps in close contact with the work.

• Batch control

Batch control is another important, type of production control which is frequently found in the foodprocessing industries. Thus, production control in batch-system of control operates with a set ofingredients that are proportionally related and handled one batch at a time.

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Production planning and control in continuous-production systems.

Production systems may be continuous or intermittent. The continuous production systems are characterised by:

� Fixed-path material handling equipment,

� High volume of production,

� Product layouts,

� Production of standardised products,

� Production to stock or long-range orders,

� The use of special-purpose machines or automation.

Production planning and control in continuous-production systems involve two activities:

� Assuring that supply of raw materials and supplies are on hand to keep the production system suppliedand assuring that finished products are moved from the production-system,

� Maintaining a constant rate of flow of the production, so that the system can operate near capacity insome case or can meet the quantity requirements of the production.

Production planning in intermittent production systems:

The intermittent production systems are characterized by the following:

� General purpose production machines are normally utilised and process layout is favoured.

� Materials handling equipment is typically of the varied path type such as hand trucks and forklifttrucks.

� Relatively high cost, skilled labour is needed to turn out the various quantities and types of products.

� The company generally manufactures a wide variety of products; for the majority of items, salesvolumes and consequently production order sizes are small in relation to the total production.


Problem: 1

Machines K and L, both capable of manufacturing an industrial product, compare as follows:

Machine K Machine L

Investment Rs. 60,000 Rs. 1,00,000

Interest on borrowed capital 15% 15%

Operating cost (wages, power, etc.) per hour Rs. 12 Rs. 10

Production per hour 6 pieces 10 pieces

The factory whose overhead costs are Rs. 1,20,000 works effectively for 4,000 hours in 2 shifts duringthe year. (i) Justify with appropriate calculations which of the two machines you would choose forregular production. (ii) If only 4000 pieces are to be produced in a year, which machine would give thelower cost per piece. (iii) For how many pieces of production per year would the cost of production be


65

same on either machine? (For above comparisons, the cost of material may be excluded as being thesame on both machines.)

Solution:

Machine K Machine L

Annual interest charges (Rs. 60,000 x 15) / 100 (Rs. 1,00,000 x 15) / 100

(fixed cost) = Rs. 9,000 = Rs. 15,000

Annual operating charges 4,000 x 12 4,000 x 10

= Rs. 48,000 = Rs. 40,000

Total annual charges = Rs. 57,000 = Rs. 55,000

Annual output = 4,000 x 6 = 24,000 = 4,000 x 10 = 40,000

Cost per unit = 5,700 / 2,400 = 55,000 / 40,000

= Rs. 2.375 = Rs. 1.375

(i) Thus machine L should be chosen for regular production.

(ii) If only 4,000 pieces are to be produced in a year

Interest cost Rs. 9,000 Rs. 15,000

Operating cost (4,000 / 6) x 12 = Rs. 8,000 (4,000 / 10) x 10 = Rs. 4,000

Total cost = Rs. 17,000 = Rs. 19,000

Cost per unit (17,000 / 4,000) = Rs. 4.25 (19,000 / 4,000) = Rs. 4.75

Thus, machine K gives the lower cost per piece.

(iii) Interest charge = Rs. 9,000 = Rs. 15,000

Operating cost per piece = 12 / 6 = 2 = 10/10 = Re. 1

Let the production be = X units,

Then 2X + 9,000 = X + 15,000 or, 2X – X = 15,000 – 9,000 or, X = 6,000 pieces.

For 6,000 piece of production per year the cost of production will be the same (Rs. 21,000) on eithermachine.

Problem: 2

A department of a company has to process a large number of components/month. The process equipmenttime required is 36 minutes/component, whereas the requirement of an imported process chemical is 1.2litres/component. The manual skilled manpower required is 12 minutes/component for polishing andcleaning. The following additional data is available:

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Availability/month Efficiency of utilisation

Equipment hour 500 85%

Imported chemicals 1000 95%

Skilled manpower - hours 250 65%

(i) What is the maximum possible production under the current conditions? (ii) If skilled man-power availability is increased by overtime by 20%, what will be the impact on productionincrease?

Solution:

(i) Actual Equipment Hrs. used = 500 × 85/100 = 425 Hrs.

Possible output = 425 × (60/36) = 708 Components

(ii) Imported chemicals = 1,000 × 95/100 = 950 litres, actually used;

Possible output = 950/1.2 = 792 Components

(iii) Skilled manpower Hrs. used = 250 × 65/100 = 162.5 Hrs.

Possible output =162.5 × (60/12) = 813 Components

The bottleneck capacity = 708 Components.

(1) Maximum possible production under the given conditions = 708 Components.

(2) There will be no impact on production increase if skilled manpower is increased by overtime by20% as the bottleneck in output is equipment hours.

Problem: 3

A manufacturing enterprise has introduced a bonus system of wage payment on a slab-rate based on costof production towards labour and overheads.

The slab-rate being

1% - 10% saving in production cost 5% of saving

Between 11%-20% saving in production cost 15%



Above 70% saving in production cost 50%

The rate per hour for three workers A, B, C are Rs. 5, Rs. 5.50 and Rs. 5.25 respectively. The overheadrecovery rate is 500% of production wages and the material cost is Rs. 40 per unit. The standard cost ofproduction per unit is determined at Rs. 160 per unit.

If the time taken by A, B, C to finish 10 units is 26 hours, 30 hours and 16 hours respectively, what is theamount of bonus earned by the individual workers and actual cost of production per unit?


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Solution:

A B C

Unit produced 10 10 10

Wage rate 5.00 5.50 5.25

Time taken 26 hours 30 hours 16 hours

Wage payable 130.00 165.00 84.00

Overhead recovery 650.00 825.00 420.00

Materials 400.00 400.00 400.00

Total cost of production 1,180.00 1,390.00 904.00

Standard cost of production 1,600.00 1,600.00 1,600.00

Saving in cost of production 420.00 210.00 696.00

% of savings 26.25% 13.13% 43.50%

Bonus slab 30% 15% 40%

Bonus Amount 126.00 31.50 278.40

Actual cost of production 1,306.00 1,421.50 1,182.40

Cost/unit (Rs.) 130.60 142.15 118.24

Problem: 4

Calculate the break-even point for the following:

Production Manager of a unit wants to know from what quantity he can use automatic machine againstsemi-automatic machine.

Data Automatic Semi-automatic

Time for the job 2 mts 5 mts

Set up time 2 hrs 1.5 hrs

Cost per hour Rs. 20 Rs. 12

Solution:

Let x be the break-even quantity between automatic and semi-automatic machines. This means, forvolume of output V, the total cost of manufacture is the same on both automatic and semi-automaticmachines.

For quantity = x units

Total manufacturing cost on automatic machines = 2

2.0+ 60

⎛ ⎞⎜ ⎟⎝ ⎠

x × 20 Rs.

Total manufacturing cost on semi-automatic machines = 5

1.5+ 60

⎛ ⎞⎜ ⎟⎝ ⎠

x ×12 Rs.

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If ‘x’ is the break-even quantity, then

22.0 +

60

x⎛ ⎞⎜ ⎟⎝ ⎠ × 20 =

51.5 +

60

x⎛ ⎞⎜ ⎟⎝ ⎠ × 12

40 +2

2060

x × = 18 +5

1260

x ×

40 + 2

3x

= 18 + x

x - 2

3x

= 40 - 18 = 22

3x

= 22

x = 66 units

Hence for quantity upto 65, a semi-automatic machine will be cheaper. For quantity 66, both semi-automatic and automatic machines are equally costly. For quantity more than 66, automatic machinebecomes cheaper than semi-automatic machine.

Problem: 5

Two alternative set-ups, A and B are available for the manufacture of a component on a particular machine,where the operating cost per hour is Rs. 20/-.

Set-up A Set-up B

Components/set-up 4,000 pieces 3,000 pieces

Set-up cost Rs. 300/- Rs.1,500/-

Production rate/hour 10 pieces 15 pieces

Which of these set-ups should be used for long range and economic production?

Solution:

Considering one set-up

Set-A Set-up B

Set-up cost per year Rs. 300/- Rs. 1,500/-

Operating hours / set-up400010

= 400 hours300015

= 200 hours

Operating cost 400 x 20 = Rs. 8,000 200 x 20 = Rs. 4,000

Total manufacturing cost 300 + 8,000 = Rs. 8,300 1,500 + 4,000 = Rs. 5,500

Manufacturing cost per piece83004000

= Rs. 2.07555003000

= Rs. 1.833


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Assuming that the machine is used for production for one year having 2,000 hours of working. Forannual production,

Set-up A Set-up B

No. of set-ups2000400

= 52000200

= 10

Set-up cost per year 5 x 300 = Rs. 1,500 10 x 1,500 = Rs. 15,000

No. of units produced per year 2,000 x 10 = Rs. 20,000 2,000 x 15 = Rs. 30,000

Total annual manufacturing cost 1,500 + 40,000 = Rs. 41,500 15,000 + 40,000 = Rs. 55,000

Manufacturing cost per unit83004000

= Rs. 2.07555003000

= Rs. 1.833

Since the manufacturing cost for set B is less, use set-up B for long range and economic production.

2.2 Forecasting

Forecasting means peeping into the future. As future is unknown and is anybody’s guess but the businessleaders in the past have evolved certain systematic and scientific methods to know the future by scientificanalysis based on facts and possible consequences. Thus, this systematic method of probing the future iscalled forecasting. In this way forecasting of sales refers to an act of making prediction about future salesfollowed by a detailed analysis of facts related to future situations and forces which may affect the businessas a whole.

Foresight is not the whole of management, but at least it is an essential part of management and accordingly,to foresee in this context means both to assess the future and make provisions for it, that is forecasting isitself action already. Forecasting as a kind of future picture wherein proximate events are outlined withsome distinctness, while remote events appear progressively less distinct and it entails the running of thebusiness as foresee and provide means to run the business over a definite period.

As far as the marketing manager is concerned the sales forecast is an estimate of the amount of unit salesfor a specified future period under the proposed marketing plan or program. It may also be defined as anestimate of sales in rupees of physical units for a specified future period under a proposed marketing planor program and under an assumed set of economic and other force outside the organisation for which theforecast is made.

When we consider the function of production and operations management, no doubt Production andOperation departments will produce goods as per the sales program given by the sales department, but ithas to prepare forecast regarding machine capacity required, materials required and time required forproduction and so on. This needs the knowledge of what exactly happened in the production shop inprevious periods.

Making of a proper forecast requires the assessment of both controllable and uncontrollable factors (botheconomic and non economic) inside and outside the organisation.

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All business and industrial activities revolve around the sale and its future planning. To know what abusiness will do we must know its future sales. So, sales forecasting is the most important activity in thebusiness because all other activities depend upon the sales of the concern. Sales forecasting as a guidingfactor for a firm because it enables the firm to concentrate its efforts to produce the required quantities, atthe right time at reasonable price and of the right quality. Sales forecasting is the basis of planning thevarious activities i.e.; production activities, pricing policies, programme policies and strategies, personnelpolicies as to recruitment, transfer, promotion, training, wages etc.

The period of forecasting, that is the time range selected for forecasting depends on the purpose for whichthe forecast is made. The period may vary from one week to some years. Depending upon the period, theforecast can be termed as ‘Short range forecasting’, medium range forecasting’ and ‘Long range forecasting’.‘Short range forecasting period may be one week, two weeks or a couple of months. Medium rangeforecasting period may vary from 3 to 6 months. Long range forecasting period may vary from one year toany period. The objective of above said forecast is naturally different.

In general, short term forecasting will be of more useful in production planning. The manager who doesshort range forecast must see that they are very nearer to the accuracy.

In long range forecast, the normal period used is generally 5 years. In some cases it may extends to 10 to 15years also. The purpose of long range forecast is:

(i) To work out expected capital expenditure for future developments or to acquire new facilities,

(ii) To determine expected cash flow from sales,

(iii) To plan for future manpower requirements,

(iv) To plan for material requirement,

(v) To plan for Research and Development. Here much importance is given to long range growth factor.

In case of medium range forecasting the period may extend over to one or two years. The purpose of thistype of forecasting is:

(i) To determine budgetary control over expenses,

(ii) To determine dividend policy,

(iii) To find and control maintenance expenses,

(iv) To determine schedule of operations,

(v) To plan for capacity adjustments.

In case of short-term forecast, which extends from few weeks to three or six months and the followingpurposes are generally served:

(i) To estimate the inventory requirement,

(ii) To provide transport facilities for finished goods,

(iii) To decide work loads for men and machines,

(iv) To find the working capital needed,

(v) To set-up of production run for the products,


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(vi) To fix sales quota,

(vii) To find the required overtime to meet the delivery promises.

Everyone who use the forecast for one purpose or the other expects that they need that forecast should beaccurate. But it is practically impossible to forecast accurately. But decisions are made everyday to run thebusiness by using the best information available with them. Management scientists have developed variousmethods for forecasting. One has to decide which method has to be used to suit the information availablewith him and to suit his needs. The manager, who is concerned with forecasting, must have knowledge offactors influencing forecast. Various factors that influence the forecast are:

(i) Environmental changes,

(ii) Changes in the preference of the user,

(iii) Number of competitive products,

(iv) Disposable income of the consumer.

In forecasting the production important factors to be considered are:

(i) Demand from the marketing department,

(ii) Rate of labours absenteeism,

(iii) Availability of materials,

(iv) Available capacity of machines,

(v) Maintenance schedules,

(vi) Delivery date schedules.

Steps in forecasting

Whatever may be the method used for forecasting, the following steps are followed in forecasting.

(a) Determine the objective of forecast: What for you are making forecast? Is it for predicting the demand?Is it to know the consumer’s preferences? Is it to study the trend? You have to spell out clearly the useof forecast.

(b) Select the period over which the forecast will be made? Is it long-term forecast or medium-term forecastor short-term forecast? What are your information needs over that period?

(c) Select the method you want to use for making the forecast. This method depends on the period selectedfor the forecast and the information or data available on hand. It also depends on what you expectfrom the information you get from the forecast. Select appropriate method for making forecast.

(d) Gather information to be used in the forecast. The data you use for making forecasting to produce theresult, which is of great use to you. The data may be collected by:

(i) Primary source: This data we will get from the records of the firm itself.

(ii) Secondary source: This is available from outside means, such as published data, magazines,educational institutions etc.

(e) Make the forecast: Using the data collected in the selected method of forecasting, the forecast is made.

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Forecasting Methods:

Methods or techniques of sales forecasting: Different authorities on marketing and production have devisedseveral methods or techniques of sales or demand forecasting. The sales forecasts may be result of whatmarket people or buyers say about the product or they may be the result of statistical and quantitativetechniques. The most common methods of sales forecasting are:

1. Survey of buyer’s inventions or the user’s expectation method: Under this system of sales forecastingactual users of the product of the concern are contacted directly and they are asked about their intentionto buy the company’s products in an expected given future usually a year. Total sales forecasts of theproduct then estimated on the basis of advice and willingness of various customers. This is mostdirect method of sales forecasting.

The chief advantages of this method are:

(i) Sales forecast under this method is based on information received or collected from the actualusers whose buying actions will really decide the future demand. So, the estimates are correct.

(ii) It provides a subjective feel of the market and of the thinking behind the buying intention of theactual uses. It may help the development of a new product in the market.

(iii) This method is more appropriate where users of the product are numbered and a new product isto be introduced for which no previous records can be made available.

(iv) It is most suitable for short-run forecasting.

2. Collective opinion or sales force composite method: Under this method, views of salesmen, branchmanager, area manager and sales manager are secured for the different segments of the market.Salesmen, being close to actual users are required to estimate expected sales in their respective territoriesand sections. The estimates of individual salesmen are then consolidated to find out the total estimatedsales for the coming session. These estimates are then further examined by the successive executivelevels in the light of various factors like proposed changes in product design, advertising and sellingprices, competition etc. before they are finally emerged for forecasting.

3. Group executive judgement or executive judgement method: This is a process of combining, averagingor evaluating, in some other way, the opinions and views of top executives. Opinions are sought fromthe executives of different fields i.e., marketing; finance; production etc. and forecasts are made.

4. Experts’ opinions: Under this method, the organisation collects opinions from specialists in the fieldoutside the organisation. Opinions of experts given in the newspapers and journals for the trade,wholesalers and distributors for company’s products, agencies or professional experts are taken. Byanalysing these opinions and views of experts, deductions are made for the company’s sales, andsales forecasts are done.

5. Market test method: Under this method seller sells his product in a part of the market for sometimesand makes the assessment of sales for the full market on the bases of results of test sales. This methodis quite appropriate when the product is quite new in the market or good estimators are not availableor where buyers do not prepare their purchase plan.

6. Trend projection method: Under this method, a trend of company’s or industry’s sales is fixed withthe help of historical data relating to sales which are collected, observed or recorded at successive


73

intervals of time. Such data is generally referred to as time series. The change in values of sales isfound out. The study may show that the sales sometimes are increasing and sometimes decreasing,but a general trend in the long run will be either upward or downward. It cannot be both ways. Thistrend is called secular trend. The sales forecasts with the help of this method are made on the assumptionthat the same trend will continue in the future. The method which is generally used in fitting thetrend is the method of least squares or straight line trend method. With this method a straight linetrend is obtained. This line is called ‘line of best fit’. By using the formula of regression equation of Yon X, the future sales are projected.

Calculation of trend.

The trend can be calculated by the least square method as follows:

(i) Find time deviations (X) of each period from a certain period and then find the sum oftime deviation ( ΣX).

(ii) Square the time deviation of each period (X2) and then find the sum of squares of eachperiod (ΣX 2).

(iii) Multiply time deviations with the sales of each period individually (XY) and add theproduct of the column to find (ΣXY).

(iv) To find the trend (Y) this is equal to a + bX. The value of a and b may be determined byeither of the following two ways:

(a) Direct method. This method is applicable only when ΣX=0. To make ΣX=0, it is necessary thatthe time deviations should be calculated exactly from the mid point of the series. Then, the valuesof a and b will be calculated as follows:

a (average) = Y

n∑ and b (rate of growth) = 2

XY

Y∑∑

This method is simple and direct.

(b) Indirect method. This method is somewhat difficult. This method can be applied in both thecases where ΣX has any positive or negative values or ΣX is not equal to zero. The values of a andb are calculated by solving the following two equations:

Σ Y = na + bΣ X

Σ XY = aΣ X + bΣ X2

By calculating the values of a and b in the above manner, the sales can be forecasted for any futureperiod by applying the formula Y = a + bX.

7. Moving average method: This is another statistical method to calculate the trend through movingaverages. It can be calculated as follows:

An appropriate period is to be determined for which the moving average is calculated. Whiledetermining the period for moving averages, the normal cycle time of changes in the values of seriesshould be considered so that short-term fluctuations are eliminated. As far as possible, the period formoving averages should be in odd numbers such as period of 3, 5 or 7 years. The period in even

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numbers will create a problem in centralising the values of averages. The calculated values of movingaverages present the basis for determining the expected amount of sale.

8. Criteria of a good forecasting method: It cannot be said which method of sales forecasting is the bestbecause everyone has merits and demerits of its own. The suitability of a method depends on variousfactors such as nature of theproduct, available time and past records, wealth and energy, degree ofaccuracy and the forecaster etc. of an enterprise. However, in general, a good forecasting methodmust possess the following qualifications.

(i) Accuracy: Accuracy of the forecasting figures is the life blood of the business because manyimportant plans and programmes, policies andstrategies are prepared and followed on the basisof such estimates. If sales forecasts are wrong, the businessman suffer a big loss. Hence, themethod of forecasting to be applied must amount to maximum accuracy.

(ii) Simplicity: The method for forecasting should be very simple. If the method is difficult or technical,then there is every possibility of mistake. Some information are collected from outside and thatwill remain unanswered or inaccurate replies will be received, if the method is difficult.Management must also be able to understand and have confidence in the method.

(iii) Economy: The method to be used should be economical taking into account the importance ofthe accuracy of forecast. Costs must be weighed against the importance of the forecast to theoperations of the business.

(iv) Availability: The method should be such for which the relevant information may be availableimmediately with reasonable accuracy. Moreover, the technique must give quick results anduseful information to the management.

(v) Stability: The data of forecasting should be such wherein the future changes are expected to beminimum and are reliable for future planning for sometime.

(vi) Utility: The forecasting technique must be easily understandable and suitable to the management.


Problem: 1

An investigation into the demand for colour TV sets in 5 towns has resulted in the following data:

Population of the town (in lakhs) X: 5 7 8 11 14

No of TV sets demanded (in thousands) Y: 9 13 11 15 19

Fit a linear regression of Y on X and estimate the demand for CTV sets for two towns with a populationof 10 lakhs and 20 lakhs.


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Solution:

Computation of trend values

Population Sales of CTV Squares of the Product of population and(in lakhs) (in thousands) population sales of colour TV

X Y X² XY

5 9 25 45

7 13 49 91

8 11 64 88

11 15 121 165

14 19 196 266

ΣX = 45 Σy = 67 ΣX² = 455 ΣXY = 655

Regression equation of Y on X

Y = a + bX

To find the values of a and b, the following two equations are to be solved

ΣY = na + bΣX ... (i)

ΣXY = aΣX + bΣX2 ... (ii)

By putting the values we get

67 = 5a + 45b ... (iii)

655 = 45a + 455b ... (iv)

Multiplying equation (iii) by 9 and putting it as no. (v) we get,

603 = 45a + 405b ... (v)

By deducting equation (v) from equation (iv); we get 52 = 50b

b = 5250

= 1.04

By putting the value of b in equation (iii), we get

67 = 5a + 45 × 1.04

or, 67 = 5a + 46.80

or, 67-46.80 = 5a

or, 5a = 20.20

or, a =20.20

5

or a = 4.04

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Now by putting the values of a, b and X (10 lakhs) in regression equation of Y on X, we get,

Y = a + bX

or, Y = 4.04 + 1.04 (10)

or, Y = 4.04 + 10.40 or 14.44 thousand CTV sets.

Similarly sales estimates for town having population of 20 lakhs, by putting the values of X, a and b inregression equation can be found as

Y = 4.04 + 1.04 (20)

= 4.04 + 20.80 = 24.84 thousands CTV sets.

Hence expected demand for CTV for two towns will be 14.44 thousand and 24.84 thousand CTV sets.

Problem: 2

The annual sales of truck tyres manufactured by a company are as follows—

Year (X) 2002 2003 2004 2005 2006

Sales (‘000 units) (Y) 53 64 86 54 83

Fit a linear trend equation to the sales figures and estimate the sales for 2007.

Solution:

Computation of Trend Values

Years Time Deviation Sales in Squares of Product of timefrom 2004 (‘000 units) time dev. deviations and sales

X Y mX² XY

2002 –2 53 4 –106

2003 –1 64 1 –64

2004 0 86 0 0

2005 +1 54 1 +54

2006 +2 83 4 +166

n = 5 ΣX = 0 ΣY = 340 ΣX² = 10 ΣXY = + 50

Regression equation of Y on X—

Y = a + bXFor calculating the values of a and b

a = ∑Y

n =

3405

or 68

b = 2

5010

=∑∑

XY

X = 5


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Hence, regression equation comes to Y = 68 + 5X with the help of this equation, the trend value for2007 can be calculated as follows—

Y2007 = 68 + 5(5) = 68 + 25 = 93

The estimated sales for 2007 will be 93,000 units.

Problem: 3.

From the following time series data of sale project the sales for the next three years.

Year 2001 2002 2003 2004 2005 2006 2007

Sales (`000 units) 80 90 92 83 94 99 92

Solution.

Computation of Trend Values

Years Time Deviation Sales in Squares of Product of timefrom 2004 (`000 units) time dev. deviations and sales

X Y X² XY

2001 –3 80 9 –240

2002 –2 90 4 –180

2003 –1 92 1 –92

2004 0 83 0 0

2005 +1 94 1 +94

2006 +2 99 4 +198

2007 +3 92 9 +276

n = 5 ΣX = 0 ΣY = 630 ΣX² = 28 ΣXY = + 56


Y = a + bX

To find the values of a and b

a = ∑Y

n =

6307

= 90

b = 2

XY

X∑∑

= 5628

= 2

Hence regression equation comes to Y = 90 + 2X. With the help of this equation we can project thetrend values for the next three years, i.e. 2008, 2009 and 2010.

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Y2008 = 90 + 2(4) = 90 + 8 = 98 (000) units.

Y2009 = 90 + 2(5) = 90 + 10 = 100 (000) units.

Y2010 = 90 + 2(6) = 90+ 12= 102 (000) units.

Problem: 4

Project the trend of sales for the next 5 years from the following data –

Year 2003 2004 2005 2006 2007

Sales (‘000units) 120 140 120 150 170

Solution.

Calculation of trend values of sales

Years Sales Time deviation Squares of Product of time(in lakh of Rs.) (from 2005) time deviation deviations and sales

Y X X² XY

2003 120 –2 4 –240

2004 140 –1 1 –140

2005 120 0 0 0

2006 150 +1 1 150

2007 170 +2 4 340

n = 5 ΣY = 700 ΣX = 0 ΣX² = 10 ΣXY = 110


Y = a + bX

To find values of a and b

a = Y

n∑ =

7005

= 140

b = 2

XY

X∑∑

= 11010

= 11

Hence regression equation is a + bX or 140 + 11X. With the help of this equation we can project thetrend for the next five years as follows:

Y2008 = 140 + 11 × 3 = 140 + 33 = 173 lakh rupees.

Y2009 = 140 + 11 × 4 = 140 + 44 = 184 lakh rupees.

Y2010 = 140 + 11 × 5 = 140 + 55 = 195 lakh rupees.

Y2011 = 140 + 11 × 6 = 140 + 66 = 206 lakh rupees.

Y2012 = 140 + 11 × 7= 140 + 77 = 217 lakh rupees.


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Problem: 5

An investigation into the use of scooters in 5 towns has resulted in the following data:

Population in town

Population in town (X) 4 6 7 10 13(in lakhs)

No. of scooters (Y) 4,400 6,600 5,700 8,000 10,300

Fit a linear regression of Y on X and estimate the number of scooters to be found in a town with apopulation of 16 lakhs.

Solution:

Computation of trend value

Population No. of scooters Squares of Product of population and(in lakhs) demanded population No. of scooters demanded

X Y X² XY

4 4,400 16 17,600

6 6,600 436 39,600

7 5,700 49 39,900

10 8,000 100 80,000

13 10,300 169 1,33,900

ΣX = 40 ΣY = 35,000 ΣX² = 370 ΣXY = 3,11,000

Regression equation of Y on XY = a + bXTo find the values of a and b we will have to solve the following two equations

ΣY = na + bΣX ... (i)

ΣXY = aΣX + bΣX2 ....(ii)

By putting the values, we get

35,000 = 5a + 40b ... (iii)

3,11,000 = 40a + 370b ... (iv)

By multiplying equation no. (iii) by 8 putting as equation (v) we get,

2,80,000 = 40a + 320b ... (v)

By subtracting equation (v) from equation (iv), we get

31,000 = 50b

or, 50b = 31,000

or, b =31000

50 = 620

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By substituting the value of b in equation no. (iii), we get

35,000 = 5a + 40b

or 35,000 = 5a + 40 × 620

or 35,000 = 5a + 24,800

or 10,200 = 5a

or a =10200

5 = 2040

Now putting the value of a, b and X (16 lakhs) in regression equation of Y on X, we

get Y = a + bX

or, Y = 2040 + 620 (16)

or Y = 2040 + 9920

or Y = 11,960

Hence, the expected demand of scooters for a town with a population of 16 lakhs will be 11,960scooters.

Problem 6.

An investigation into the demand for TV sets in 7 towns has resulted in the following data:

Population (m 000) X : 11 14 14 17 17 21 25

No. of TV sets demande Y : 15 27 27 30 34 38 46

Fit a linear regression of Y on X, and estimate the demand for TV sets for a town with a population of30,000.

Solution

Population No. of TV sets Squares of Product of population and(in ‘000) demanded population No. TV sets demanded

X Y X² XY

11 15 121 165

14 27 196 378

14 27 196 378

17 30 289 510

17 34 289 578

21 38 441 798

25 46 625 1150

ΣX = 119 ΣY = 217 ΣX² = 2157 ΣXY = 3957n = 7


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Regression equation of Y on X:Y = a + bX

To find the value of a and b the equations are to be solved:

ΣY = na + bΣX ... (i)



217 = 7a + 119b ...(iii)

3957 = 119a + 2157b ...(iv)

Multiplying equation no. (iii) by 17 and putting it as no. (v) we get

3689 = 119a + 2023b ...(v)

By deducting equation (v) from (iv)

we get 268 = 134b

or 134b = 268

or, b =268134

= 2

By substituting the value of b in equation no. (iii), we get

217 = 7a + 119 × 2

or 7a + 238 = 217

or 7a = 217 - 238 = -21 or a = -3

Now, by putting the values of a, b and X (i.e.,) in regression equation of Y on X, we get

Y = -3 + 2 × 30 = -3 + 60 = 57

Hence, the expected demand for TV sets for a town with a population of 30,000 will be 57 sets.

Problem 7:

An investigation into the demand for coolers in 5 towns has resulted in the following data:

Population of the town X : 5 7 8 11 14(in lakhs)

No. of coolers demanded Y : 45 65 55 75 95

Fit a linear regression of Y on X and estimate the demand for coolers for a town with a population of25 lakhs.

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Solution:

Computation of demand for coolers for a population of 20 lakhs

Towns Population of No. of coolers Squares of Production and no.town (in lakhs) demanded population of coolers demanded

X Y X² XY

A 5 45 25 225

B 7 65 49 455

C 8 55 64 440

D 11 75 121 825

E 14 95 196 1,330

n = 5 ΣX = 45 ΣY = 335 ΣX² = 455 ΣXY = 3,275

Regression equation of Y on X.

Y = a + bX

To find the values of a and b the following two regression equations are to be solved :

ΣY = na + bΣX .....(i)

ΣXY = aΣX + bΣX2 .....(ii)


335 = 5a + 45b .... (iii)

3,275 = 45a + 455b .... (iv)

By multiplying equation (iii) by 9, we get

3,015 = 45a + 405b .... (v)

By subtracting equation (v) from (iv) we get

45a + 455b = 3,275

45a + 405b = 3,015

50b = 260

or b = 260/50 = 5.2

By putting the value of b in equation (iii), we get

335 = 5a + 45 × 5.2

or 5a = 335 - 234 = 101

or a = 101/5 = 20.2

By putting the value of a, b and X (which is 20) in regression equation of Y on X, we get

Y = a + bX


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Y = 20.5 + 5.2 (20)

Y = 20.2 + 104 = 124.2 or 124

or say expected demand for room coolers for a town having a population of 20 lakhs will be 124 roomcoolers.

Problem: 8

An investigation into the demand for coolers in five towns has resulted in the following data:

Population of the town X : 4 6 7 10 13(in lakhs)

No. of coolers demanded Y : 40 60 50 70 90

Fit a linear regression of Y on X and estimate the demand for coolers for a town with a population of20 lakhs.

Solution:

Computation of trend values of sales.

Towns Population Demand for Squares of Product of Population Trend values(in lakhs) room coolers population & demand XY (Y = a + bX)

X Y X² XY Y

A 4 40 16 160 20.40 + (5.2 x 4) = 41.2

B 6 60 36 360 20.4 + (5.2 x 6) = 51.6

C 7 50 49 350 20.4 + (5.2 x 7) = 56.8

D 10 70 100 700 20.4 + (5.2 x 10) = 72.4

E 13 90 169 1,170 20.4 + (5.2 x 13) = 88.0

n = 5 ΣX = 40 ΣY = 310 ΣX² = 370 ΣXY = 2,740 = 310

Regression equation of Y on X = Y = a + bX

To find out the value of a and b, the following two regression equations are to be solved:

ΣY = na + bΣX ... (i)


By putting the values in the above two equations

310 = 5a + 40b ...(iii)

2,740 = 40a + 370b ... (iv)

By multiplying the equation (iii) with 8 and deducting it from equation (iv)

2,480 = 40a + 320b ... (v)

2,740 = 40a + 370b ... (vi)

-260 = - 50b

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or 50b = 260

or b = 260/50 = 5.2

By substituting the value of b in equation (iii)

310 = 5a + 40 × 5.2

or 5a = 310-208 = 102

or a = 102/5 = 20.4

Now by putting the values of a and b in regression equation of Y on X, we find the following equation:

Y = 20.4 + 5.2 (20) = 20.4 + 104 = 124.4 or say 124

Problem: 9

With the help of following data project the trend of sales for the next five years:

Years 2002 2003 2004 2005 2006 2007

Sales (in lakhs) 100 110 115 120 135 140

Solution:

Computation of trend values of sales

Year Time deviations from Sales Squares of Product of timethe middle of 2004 and (in lakh Rs.) time deviation deviation and sales

2005 assuring 5 years = 1X Y X2 XY

2002 -5 100 25 -500

2003 -3 110 9 -330

2004 -1 115 1 -115

2005 +1 120 1 +120

2006 +3 135 9 +405

2007 + 5 140 25 +700

n = 6 ΣX = 0 ΣY = 720 ΣX² = 70 ΣXY = 280

Regression equation of Y on X:

Y = a + bX

To find the values of a and b

a = ∑Y

n =

7206

= 120


85

b = 2

∑∑

XY

X = 28070

= 4

Sales forecast for the next years, i.e., 2008 to 2012

Y2008= 120 + 4 (+7) = 120 + 28 = Rs. 148 lakhs

Y2009 =120 + 4 (+9) = 120 + 36 = Rs. 156 lakhs

Y2010 = 120 + 4 (+11) = 120 + 44 = Rs. 164 lakhs.

Y2011 =120 + 4 (+13) = 120 + 52 = Rs. 172 lakhs.

Y2012 = 120 + 4 (+15) = 120 + 60 = Rs. 180 lakhs.

Problem: 10

There exists a relationship between expenditure on research and its annual profit. The details of theexpenditure for the last six years is given below. Estimate the profit when the expenditure is 6 units

Year Expenditure for research Annual Profit

(x) (y)

2001 2 20

2002 3 25

2003 5 34

2004 4 30

2005 11 40

2006 5 31

2007 6 ?

(One unit corresponds to 1 Crore Rs.)

Solution:

Year Expenditure for Annual Profit xy x²research (x)

2001 2 20 40 4

2002 3 25 75 9

2003 5 34 170 25

2004 4 30 120 16

2005 11 40 440 121

2006 5 31 155 25

Total 30 180 1000 200

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x =306

= 5 and y = 180

6 = 30

The values a and b are computed as follows: for a linear regression equation

y = a + bX

b = 2 2

−

−∑∑

xy n x y

x n x

b =1000 6 5 30

200 6 5 5

− × ×− × × =

1000 900

200 150

−− = 2

a = y – b x = 30 - 2 x 5 = 20

Thus, the model is y = 20 + 2x.

The profit when the expenditure is 6 units is

y = 20 + 2 × 6 = 32 units of Rs.

2.3 Capacity Planning and Utilization

Capacity Planning:

The effective management of capacity is the most important responsibility of production and operationsmanagement. The objective of capacity management i.e., planning and control of capacity is to match thelevel of operations to the level of demand.

Capacity planning is concerned with finding answers to the basic questions regarding capacity such as:

(i) What kind of capacity is needed?

(ii) How much capacity is needed?

(iii) When this capacity is needed?

Capacity planning is to be carried out keeping in mind future growth and expansion plans, market trends,sales forecasting, etc. Capacity is the rate of productive capability of a facility. Capacity is usually expressedas volume of output per period of time.

Capacity planning is required for the following:

• Sufficient capacity is required to meet the customers demand in time,

• Capacity affects the cost efficiency of operations,

• Capacity affects the scheduling system,

• Capacity creation requires an investment,

• Capacity planning is the first step when an organisation decides to produce more or new products.

Capacity planning is mainly of two types:

(i) Long-term capacity plans which are concerned with investments in new facilities and equipments.These plans cover a time horizon of more than two years.

(ii) Short-term capacity plans which takes into account work-force size, overtime budgets, inventoriesetc.


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Capacity refers to the maximum load an operating unit can handle. The operating unit might be a plant, adepartment, a machine, a store or a worker. Capacity of a plant is the maximum rate of output (goods orservices) the plant can produce.

The production capacity of a facility or a firm is the maximum rate of production the facility or the firm iscapable of producing. It is usually expressed as volume of output per period of time (i.e., hour, day, week,month, quarter etc.). Capacity indicates the ability of a firm to meet market demand - both current andfuture.

Effective Capacity can be determined by the following factors:

Facilities - design, location, layout and environment.

Product - Product design and product-mix.

Process - Quantity and quality capabilities.

Human factors - Job content, Job design, motivation, compensation, training and experience of labour,learning rates and absenteeism and labour turn over.

Operational factors - Scheduling, materials management, quality assurance, maintenance policies, andequipment break-downs.

External factors - Product standards, safety regulations, union attitudes, pollution control standards.

Measurement of capacity

Capacity of a plant is usually expressed as the rate of output, i.e., in terms of units produced per period oftime (i.e., hour, shift, day, week, month etc.). But when firms are producing different types of products, itis difficult to use volume of output of each product to express the capacity of the firm. In such cases,capacity of the firm is expressed in terms of money value (production value) of the various productsproduced put together.

Capacity Planning Decisions

Capacity planning involves activities such as:

(i) Assessing the capacity of existing facilities.

(ii) Forecasting the long-range future capacity needs.

(iii) Identifying and analysing sources of capacity for future needs.

(iv) Evaluating the alternative sources of capacity based on financial, technological and economicalconsiderations.

(v) Selecting a capacity alternative most suited to achieve strategic mission of the firm.

Capacity planning is necessary when an organisation decides to increase its production or introduce newproducts into the market or to increase the volume of production to gain the advantages of economies ofscale. Once the existing capacity is evaluated and a need for new or expanded facilities is determined,decisions regarding the facility location and process technology selection are undertaken.

When the long-range capacity needs are estimated through long-range forecasts for products, a firm mayfind itself in one of the two following situations:

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(i) A capacity shortage situation where present capacity is not enough to meet the forecast demand forthe product.

(ii) An excess or surplus capacity situation where the present capacity exceeds the expected future demand.

Factors affecting determination of plant capacity

(i) Capital investment required,

(ii) Changes in product design, process design, market conditions and product life cycles,

(iii) Flexibility for capacity additions,

(iv) Level of automation desired,

(v) Market demand for the product,

(vi) Product obsolescence and technology obsolescence and

(vii) Type of technology selected.

Forms of capacity planning:

Based on time-horizon

(i) Long-term capacity planning and

(ii) Short-term capacity planning

Based on amount of resources employed

(i) Finite capacity planning and

(ii) Infinite capacity planning

Factors Affecting Capacity Planning: Two kinds of factors affecting capacity planning are:

(i) Controllable Factors: amount of labour employed, facilities installed, machines, tooling, shifts of workper day, days worked per week, overtime work, subcontracting, preventive maintenance and numberof production set ups.

(ii) Less Controllable Factors: absenteeism, labour performance, machine break-downs, material shortages,scrap and rework, strike, lock-out, fire accidents etc.

Capacity Requirement Planning : Capacity requirement planning (CRP) is a technique which determineswhat equipment and labour/personnel capacities are required to meet the production objectives (i.e., volumeof products) as per the master production schedule and material requirement planning (MRP-I).

Capacity Requirement Planning Strategies:

Two types of capacity planning strategies used are:

(i) “Level capacity” plan and

(ii) “Matching capacity with demand” plan.

Level capacity plan is based in “produce-to-stock and sell” approaches wherein the production systemsare operated at uniform production levels and finished goods inventories rise and fall depending uponwhether production level exceeds demand or vice versa from time period to time period (say every quarteror every month).


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“Matching capacity with demand” Plan: In this plan, production capacity is matched with the demand ineach period (weekly, monthly or quarterly demand). Usually, material flows and machine capacity arechanged from quarter to quarter to match the demand. The main advantages are low levels of finishedgoods inventory resulting in lesser inventory carrying costs. Also, the back-ordering cost is also reduced.The disadvantages are high labour and material costs because of frequent changes in workforce (hiring,training and lay-off costs, overtime or idle time cost or subcontracting costs).

Optimum Plant Capacity: Plant capacity has a great influence on cost of production with increasing volumeof production, economies of scale arises which results in reduction in average cost per unit produced.

For a given production facility, there is an optimum volume of output per year that results in the leastaverage unit cost. This level of output is called the “best operating level” of the plant.

As the volume of output increases outward from zero in a particular production facility, average unit costsfall. These declining costs are because of the following reasons: (i) Fixed costs are spread over more unitsproduced, (ii) Plant construction costs are less, (iii) Reduced costs of purchased material due to quantitydiscounts for higher volume of materials purchased and (iv) Cost advantages in mass production processes.Longer production runs (i.e., higher batch quantity of products produced) have lesser setup cost per unitof product produced, lesser scrap etc., resulting in savings which will reduce the cost of production perunit. This is referred to as “economies of scale”. But this reduction in per unit cost will be only upto certainvolume of production. Additional volumes of outputs beyond this volume results in ever-increasing averageunit production cost. This increase in cost per unit arise from increased congestion of materials and workers,which decreases efficiency of production, and due to other factors such as difficulty in scheduling, damagedproducts, reduced employee morale due to excessive work pressure, increased use of overtime etc., resultingin “diseconomies of scale”. Hence, the plant capacity should be such that the optimum level of productionwhich gives the minimum average cost of production per unit should be possible. This plant capacity isreferred to as optimum plant capacity.

Balancing the Capacity: In firms manufacturing many products (a product line or a product-mix) the loadon different machines and equipments vary due to changes in product-mix. When the output rates ofdifferent machines do not match with the required output rate for the products to be produced, there willbe an imbalance between the work loads of different machines. This will result in some machine or equipmentbecoming a “bottleneck work centre” thereby limiting the plant capacity which wills in-turn increase theproduction costs per unit.

To overcome problem of imbalance between different machines, additional machines or equipments areadded to the bottleneck work-centre to increase the capacity of the bottle-neck work centre to match withthe capacity of other work centres. Adding new machines or equipments to bottleneck work centres toremove the imbalance in capacity between various work centres is found to be economical than givingexcessive overtime to workers working in bottle-neck centres which increases production costs. Anothermethod to remove imbalance is to subcontract excess work load of bottleneck centres to outside vendors orsubcontractors. Another way to balance capacities is to try to change the productmix by manipulating thesales for different products to arrive at a suitable product-mix which loads all work centres almost uniformly.

Implications of Plant Capacity

There are two major cost implications of plant capacity:

(i) Changes in output of an existing plant of certain installed capacity affect the production costs.

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(ii) Changes in the plant capacity by changing the size of a plant have significant effects on costs.

Factors influencing Effective Capacity

The effective capacity is influenced by – (1) Forecasts of demand, (2) Plant and labour efficiency, (3)Subcontracting, (4) Multiple shift operation, (5) Management policies.

Forecasts of demand: Demand forecast is going to influence the capacity plan in asignificant way. As such, it is very difficult to forecast the demand with accuracy as it changes significantlywith the product life-cycle stage, number of products. Products with long lifecycle usually exhibit steadydemand growth compared to one with shorter life-cycle. Thus the accuracy of forecast influences the capacityplanning.

Plant and labour efficiency: It is difficult to attain 100 per cent efficiency of plant and equipment. Theefficiency is less than 100 percent because of the enforced idle time due to machine breakdown, delays dueto scheduling and other reasons. The plant efficiency varies from equipment to equipment and fromorganisation to organisation. Labour efficiency contributes to the overall capacity utilisation. The standardtime set by industrial engineer is for a representative or normal worker. But the actual workers differ intheir speed and efficiency. The actual efficiency of the labour should be considered for calculating efficiency.Thus plant and labour efficiency are very much essential to arrive at realistic capacity planning.

Subcontracting: Subcontracting refers to off loading, some of the jobs to outside vendors thus hiring thecapacity to meet the requirements of the organisation. A careful analysis as to whether to make or to buyshould be done. An economic comparison between cost to make the component or buy the component isto be made to take the decision.

Multiple shift operation: Multiple shifts are going to enhance the firm’s capacityutilisation. But especially in the third shift the rejection rate is higher. Specially for process industrieswhere investment is very high it is recommended to have a multiple shifts.

Management policy: The management policy with regards to subcontracting, multiplicity of shifts (decisionregarding how many shifts to operate), which work stations or departments to be run for third shift, machinereplacement policy, etc., are going to affect the capacity planning.

Factors favouring over capacity and under capacity

It is very difficult to forecast demand as always there is an uncertainty associated with the demand. Theforecasted demand will be either higher or lower than the actual demand. So always there is a risk involvedin creating capacity based on projected demand. This gives rise to either over capacity or under capacity.

The over capacity is preferred when:

(a) Fixed cost of the capacity is not very high.

(b) Subcontracting is not possible because of secrecy of design and/or quality requirement.

(c) The time required to add capacity is long.

(d) The company cannot afford to miss the delivery, and cannot afford to loose the customer.

(e) There is an economic capacity size below which it is not economical to operate the plant.

The under capacity is preferred when:

(a) The time to build capacity is short.


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(b) Shortage of products does not affect the company (i.e., lost sales can be compensated).

(c) The technology changes fast, i.e., the rate of obsolescence of plant and equipment are high.

(d) The cost of creating the capacity is prohibitively high.

Aggregate Planning:

Aggregate planning is an intermediate term planning decision. It is the process of planning the quantityand timing of output over the intermediate time horizon (3 months to one year). Within this range, thephysical facilities are assumed to be fixed for the planning period. Therefore, fluctuations in demand mustbe met by varying labour and inventory schedule. Aggregate planning seeks the best combination tominimise costs.

Production planning in the intermediate range of time is termed as ‘Aggregate Planning’. It is thus calledbecause the demand on facilities and available capacities is specified in aggregate quantities. For exampleaggregate quantities of number of Automobile vehicles, Aggregate number of soaps etc. Here the totalexpected demand is specified without regard to the product mix that makes up the specified figure.

While dealing with production problems, the planning process is normally divided in three categories.

(i) Long range Planning which deals with strategic decisions such as purchase of facilities, introductionof new products, processes etc.

(ii) Short term planning which deals with day-to-day work, scheduling and sometimes inventory problems.

(iii) Intermediate Planning or Aggregate Planning, which is in between long range and short term planning,which is concerned in generally acceptable planning taking the load on hand and the facilities availableinto considerations. In aggregate planning the management formulates a general strategy by whichcapacity can be made to satisfy demand in a most economical way during a specific moderate timeperiod, say for one year. The aggregate planning is made operational through a master schedule thatgives the manufacturing schedule (Products and dates of manufacture). Generally, day-to-day schedulesare prepared from master schedule. Facility planning and scheduling has got very close relationshipwith aggregate planning.

Aggregate Planning Strategies:

The variables of the production system are labour, materials and capital. More labour effort is required togenerate higher volume of output. Hence, the employment and use of overtime (OT) are the two relevantvariables. Materials help to regulate output. The alternatives available to the company are inventories,back ordering or subcontracting of items.

These controllable variables constitute pure strategies by which fluctuations in demand and uncertaintiesin production activities can be accommodated.

Vary the size of the workforce: Output is controlled by hiring or laying off workers in proportion to changesin demand.

Vary the hours worked: Maintain the stable workforce, but permit idle time when there is a slack andpermit overtime (OT) when demand is peak.

Vary inventory levels: Demand fluctuations can be met by large amount of inventory.

Subcontract: Upward shift in demand from low level. Constant production rates can be met by usingsubcontractors to provide extra capacity.

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Aggregate planning guidelines:

1. Determine corporate policy regarding controllable variables.

2. Use a good forecast as a basis for planning.

3. Plan in proper units of capacity.

4. Maintain the stable workforce.

5. Maintain needed control over inventories.

6. Maintain flexibility to change.

7. Respond to demand in a controlled manner.

8. Evaluate planning on a regular basis.

Properties of Aggregate Planning: To facilitate the production manager the aggregate planning musthave the following characteristics:

(i) Both out put and sales should be expressed in a logical overall unit of measuring. For example, anautomobile manufacturing can say 1000 vehicles per year, without giving the number of each verityof vehicle. Similarly a paint industry can say 10,000 litres of paint and does not mention the quantitiesof each colour.

(ii) Acceptable forecast for some reasonable planning period, say one year.

(iii) A method of identification and fixing the relevant costs associated with the plant. Availability ofalternatives for meeting the objective of the organization.

Ability to construct a model that will permit to take optimal or near optimal decisions for the sequenceof planning periods in the planning horizon.

(iv) Facilities that are considered fixed to carry out the objective.


Problem 1:

A department works on 8 hours shift, 250 days a year and has the usage data of a machine, as given below:

Product Annual Processing timedemand (units) (standard time in hours)

X 300 4.0

Y 400 6.0

Z 500 3.0

Determine the number of machines required.

Solution:

Step 1: Calculate the processing time needed in hours to produce product x, y and z in the quantitiesdemanded using the standard time data.


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Product Annual Standard Processingdemand processing needed (Hrs.)(units) per unit (Hrs.)

X 300 4.0 300 x 4 = 1200 Hrs.

Y 400 6.0 400 x 6 = 2400 Hrs.

Z 500 3.0 500 x 3 = 1500 Hrs.

Total = 5100 Hrs

Step 2 : Annual production capacity of one machine in standard hours

= 8 × 250 = 2000 hours per year

Step 3 : Number of machines required

=Work load per year

Production capacity per machine = 51002000

= 2.55 machines = 3 machines.

Problem 2:

A steel plant has a design capacity of 50,000 tons of steel per day, effective capacity of 40,000 tons of steelper day and an actual output of 36,000 tons of steel per day. Compute the efficiency of the plant and itsutilisation.

Solution:

Actual output

Efficiency of the plant = Actual output 36000

100 90%Effective Capacity 40000

⎛ ⎞= × =⎜ ⎟⎝ ⎠

Utilisation = Actual output 36000

100 72%Design Capacity 50000

⎛ ⎞ ⎛ ⎞= × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Problem 3:

An item is produced in a plant having a fixed cost of Rs. 6,000 per month, variable cost of rupees 2 per unitand a selling price of Rs. 7 per unit. Determine

(a) The break-even volume.

(b) If 1000 units are produced and sold in a month, what would be the profit?

(c) How many units should be produced to earn a profit of Rs. 4000 per month?

Solution:

(a) Break-even-volume

Fixed cost (FC) = Rs. 6000 per month

Variable cost (VC) = Rs. 2 per unit

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Selling price (SP) = Rs. 7 per unit

Let Q be the break even volume per month, then

Total cost = Fixed Cost + (Variable cost / unit) × Quantity

TC = FC + (VC ×Q) = 6000 + 2Q

Sales Revenue = Selling price per unit × Quantity = 7Q

For Q to be break-even volume,

Sales Revenue = Total cost

i.e., 7Q = 6000 + 2Q

5Q = 6000

Q =6000

5⎛ ⎞⎜ ⎟⎝ ⎠

= 1200 units / month

(b) For Q = 1000,

Profit = Sales Revenue - Total cost

= SR - (FC + VC × Q)

= (7 × 1000) - (6000 + 2 × 1000)

= (7000) - (6000 + 2000)

= Rs. 7000 - 8000 = -Rs 1000 (i.e., loss of Rs. 1000)

(c) For profit of Rs. 4000, What is Q?

SR = FC + (VC) Q + Profit

7Q = 6000 + 2Q + Profit

7Q - 2Q = Rs (6000 + 4000)

5Q = Rs. 10,000

Q =10000

5⎛ ⎞⎜ ⎟⎝ ⎠

= 2000 units

Problem 4:

A manager has to decide about the number of machines to be purchased. He has three options i.e., purchasingone, or two or three machines. The data are given below.

Number of machine Annual fixed cost Corresponding range of output

One Rs. 12,000 0 to 300

Two Rs. 15,000 301 to 600

Three Rs. 21,000 601 to 900


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Variable cost is Rs. 20 per unit and revenue is Rs. 50 per unit

(a) Determine the break-even point for each range

(b) If projected demand is between 600 and 650 units how many machines should the manager purchase?

Solution:

(i) Break-even point

Let QBEP be the break even point.

FC = Fixed cost, R = Revenue per unit, VC = Variable cost

Then QBEPR = FC + (VC) QBEP

QBEP =FC

(R – VC)

Let Q1 be the break-even-point for one machine option

Then, Q1= 12000

(50 – 20) =12000

30= 400 units

(Not within the range of 0 to 300)

Let Q2 be the break-even-point for two machines option.

Then, Q2 = 15000

(50 – 20) =15000

30 = 500 units

(within the range of 301 to 600)

Let Q3 be the break-even-point for three machines option.

Then, Q3 = 21000

(50 – 20) =21000

30 = 700 units

(with in the range of 601 to 900)

(ii) The projected demand is between 600 to 650 units.

The break even point for single machine option (i.e., 400 units) is not feasible because it exceeds therange of volume that can be produced with one machine (i.e., 0 to 300).

Also, the break even point for 3 machines is 700 units which is more than the upper limit of projecteddemand of 600 to 650 units and hence not feasible. For 2 machines option the break even volume is 500units and volume range is 301 to 600.

Hence, the demand of 600 can be met with 2 machines and profit is earned because the productionvolume of 600 is more than the break even volume of 500. If the manager wants to produce 650 unitswith 3 machines, there will be loss because the break even volume with three machines is 700 units.Hence, the manager would choose two machines and produce 600 units.

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Problem 5:

A firm has four work centres, A, B, C & D, in series with individual capacities in units per day shown in thefigure below.

Work centresActual

Raw Outputmaterials (300)

(450) (360) (340) (400)

(i) Identify the bottle neck centre.

(ii) What is the system capacity?

(iii) What is the system efficiency?

Solution:

(i) The bottle neck centre is the work centre having the minimum capacity. Hence, work centre ‘C’ isthe bottleneck centre.

(ii) System capacity is the maximum units that are possible to produce in the system as a whole.Hence, system capacity is the capacity of the bottle neck centre i.e., 340 units.

(iii) System efficiency = Actual output

System capacity

= 300340

x 100 (i.e., maximum possible output) = 88.23%

Problem 6.

A firm operates 6 days a week on single shift of 8 hours per day basis. There are 10 machines of the samecapacity in the firm. If the machines are utilised for 75 percent of the time at a system efficiency of 80percent, what is the rated output in terms of standard hours per week?

Solution

Maximum number of hours of work possible per week

= (Number of machines) × (Machine hours worked per week)

= 10 ×6 × 8 = 480 hours

If the utilisation is 75% then number of hours worked = 480 × 0.75 = 360 hours.

Rated output = utilised hours × system efficiency = 360 × 0.8 = 288 standard hours.

Problem: 7

The order position (i.e., requirements of despatch) for the next twelve months in respect of a particularproduct is as under:

A B C D


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Month Required units Month Required units

1 13,000 7 11,000

2 12,000 8 7,000

3 10,000 9 15,000

4 9,000 10 13,000

5 11,000 11 12,000

6 13,000 12 10,000

The production capacity or the shop is 10,000 units per month on regular basis and 3,000 units permonth on overtime basis. Sub-contracting can be relied upon up to a capacity of 3,000 units per monthafter giving a lead time of 3 months. Cost data reveal as under: Rs. 5.00 per piece on regular basis Rs.9.00 per piece on overtime basis Rs. 7.00 per piece on sub-contract basis Cost of carrying Inventory isRe. 1.00 per unit per month. Assuming an initial inventory of 1,000 units and that no backlogging oforders is permissible, suggest an optimal production schedule. Also work out the total cost on thebasis of the suggested schedule.

Solution:

The optimum production schedule is as follows:

Month Required No. of No. of Units Sub-contract InventoryUnits (‘000) Production (6000) (‘000 units) the end

Regular Over Order Delivered of monthTime Time placed for (‘000) units

1 13 10 2 — — 0

2 12 10 2 — — 0

3 10 10 0 3 — 0

4 9 10 0 1 — 1

5 11 10 0 — — 0

6 13 10 0 2 3 0

7 11 10 0 3 1 0

8 7 10 0 +2 — 3

9 15 10 0 — 2 0

10 13 10 0 — 3 0

11 12 10 0 — 2 0

12 10 10 0 — — 0

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Cost of production on regular basis = Rs. 1,20,000 × 5 = Rs. 6,00,000;

Cost of production on overtime = Rs. 4,000 × 9 = Rs. 36,000; Cost of sub-contracting

= Rs. 11,000 × 7 = Rs. 77,000

Cost of carrying inventory = Rs. 4,000 × 1 = Rs. 4,000; Total cost on the basis of the suggested schedule=(Rs.6,00,000 + Rs. 36,000 + Rs. 77,000 + Rs. 4,000) = Rs. 7,17,000

Problem: 8

A manufacturing company has a product line consisting of five work stations in series. The individualworkstation capacities are given. The actual output of the line is 500 units per shift.

Calculate (i) System capacity (ii) Efficiency of the production line

Workstation No. A B C D E

Capacity/Shift 600 650 650 550 600

Solution:

(i) The capacity of the system is decided by the workstation with minimum capacity/shift, i.e., thebottleneck. In the given example, the work station ‘D’ is having a capacity of 550 units/ shiftwhich is a minimum.

Therefore, the system capacity = 550 units/shift. (ii) The actual output of the line = 500 units/shift.

Therefore, the system efficiency = Actual capacitySystem capacity x 100 =

500550

x 100 = 90.91 %

Problem: 9

A company intends to buy a machine having a capacity to produce 1,70,000 good parts per annum. Themachine constitutes a part of the total product line. The system efficiency of the product line is 85%.

(i) Find the system capacity.

(ii) If the time required to produce each part is 100 seconds and the machine works for 2000 hours peryear. If the utilisation of the machine is 60% and the efficiency of the machine is 90%, compute theoutput of the machine.

(iii) Calculate the number of machines required?

Solution:

(i) System capacity = Actual output / annum

System efficiency =1,70,000

0.85 = 2,00,000 units / annum

=2,00,000

2,000 = 100 units/hours

(ii) Output per annum = Unit capacity × % utilisation × efficiency


99

Unit capacity = 60 60 sec

100 sec per unit×

= 36 units

Output per hour = 36 × 0.6 × 0.9 = 19.44 units = 20 units.

(iii) Number of machines required = System capacityOutput per hour =

10020

= 5 machines

Problem: 10

The following activities constitute a work cycle.

(i) Find the total time, theoretical output obtained from the machine.

(ii) Calculate the number of machines required to produce the three components from the informationgiven below.

Sr. No. Activity Time (min)

1. Unloading 0.25

2. Inspection 0.35

3. Loading job on machine table 0.40

4. Machine operation time 0.90

Components A B C

1. Setup time per batch 25 min 55 min 45 min

2. Operation time (min/piece) 1.75 3.0 2.1

3. Batch size 350 550 575

4. Production per month 2450 4400 2875

Solution:

(i) Total cycle time (T) = 0.25 + 0.35 + 0.40 + 0.90 = 1.90 min.

(ii) Output of the machine

Output = 601.9

= 31.5 ≈ 31 units.

(iii) Number of machines required

Assume that the plant works on the single shift basis per day of 8 hours each.

The total time required for the processing the components is given by

Total time required = Setup time + operation time.

For component A,

100


Total time required = Setup time + operation time

= Production quantity Setup time

Batch size Batch

⎡ ⎤⎛ ⎞ ⎛ ⎞×⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ + Operation time

= 2450 25 1.75

2450350 60 60

⎛ ⎞ ⎛ ⎞× + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ = 2.916+71.458=74.374 hrs.

For component B,

Total time required = 4400 55 3

4400550 60 60

⎛ ⎞ ⎛ ⎞× + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ = 22/3+220=227.33 hrs.

For component C,

Total time required = 2875 45 2.1

2875575 60 60

⎛ ⎞ ⎛ ⎞× + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ = 3.75+100.625=104.375hrs.

Total time (hrs) required to process all the three components

= 74.374 + 227.33 + 104.375 = 406.079 hrs. Total number of hours available (assuming 25 workingdays) per month = 8 × 25 = 200 hrs.

Number of machines required =Total number of machine hours required

Total number of hours available =

406.079200

= 2.030 ≈ 2 machines.

Assuming a machine efficiency of 85% and operator efficiency of 75%, the number of machines requiredare:

Total hours required per month = 406.079

0.85 0.75× = 636.98 hrs.

Number of machines required = 636.98

200 = 3.18 ≈ 4 machines.

Problem: 11

Three components are to be manufactured on three machines i.e. Center lathe, Milling machine andCylindrical grinding machine.

(i) Calculate the number of machines required of each kind to produce the components if the plant worksfor 48 hours per week.

(ii) Calculate the number of machines required assuming the machine efficiency of 75%.

(iii) How do you reduce the number of machines. The following information is given:


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Machine Component A Component B Component CSetup operation Setup operation setup operation

1 Center lathe 30 min 2min 55min 2.5 min 40 min 1.5 min

2 Milling machine 45 min 8 min 30 min 4 min – –

3 Cylindrical grinding 50 min 8 min 60 min 8 min 60 min 10 min

Other details

Lot size 350 400 600

Quantity 1750 4000 3000

demanded / month

Solution:

The total time required to process the required components on the machines

(i) Center Lathe

(a) Total time required for Component A= 1750 30 2

1750350 60 60

⎛ ⎞ ⎛ ⎞× + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ = 60.83 hours

(b) Total time required for Component B = 4000 55 2.5

4000400 60 60

⎛ ⎞ ⎛ ⎞× + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ = 175.832 hours

(c) Total time required for Component C = 3000 40 1.5

4000600 60 60

⎛ ⎞ ⎛ ⎞× + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ = 78.33 hours

Total time required to process the components on center lathe

= a +b+ c = 60.83 + 175.83 + 78.333 = 314.993 hrs/month.

Available time per machine per month = 48 × 4 = 192 hours.

Total hours required / month

No. of Lathe machines required = Total hours required/ monthTotal hours available/ month =

314.993192

= 1.64 ≈ 2 nos.

If the machine efficiency is considered as 85%, then

No. of lathes required = Total hours required/ month

Total No. of hours available / month machine efficiency×

= 314.993

192 0.75× = 2.18 ≈ 3 machines

(ii) Milling Machine

Total time required to process all the components per month

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1750 45 8 4000 30 4 1750 4000350 60 60 400 60 60

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

× + × + × + × = 508.749 hours.

Number of milling machines required

= Total hours required/ monthTotal hours available/ month =

508.749192

= 3.53 ≈ 4 nos.

Cylindrical Grinding Machines

Total time required to process all the components per month

= 1750 50 10 4000 60 8 3000 60 10

1750 4000 3000350 60 60 400 60 60 600 60 60

⎛ ⎞ ⎛ ⎞ ⎛ ⎞× + × + × + × + × + ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= [4.166 + 291.666] + [10+533.333] + [5+500] = 1344.165 hours

Number of milling machines required = Total hours required/ monthTotal hours available/ month =

1344.165 192

= 7 machines

If the machine efficiency is considered as 75%,

Number of milling machine required = 1344.165 192 0.75× = 9.33 ≈ 10 machines

(iii) Reduction in number of machines

(a) By introducing the second and third shift, the number of hours available will be increasedand hence the number of machines required will be reduced.

(b) By increasing the utilisation of the machine. The availability of the machine will be increasedby proper maintenance which reduces the break down and hence the down time. Theproduction time will be increased and hence the plant utilisation.

Problem: 12

Machines A and B are both capable of processing the product. The following informations is given

Particular Machine A Machine B

Investment Rs. 75,000 Rs. 80,000

Interest on Capital invested 10% 15%

Hourly charge (wage + power) Rs. 10 Rs. 8

Pieces produced per hour 5 8

Annual operating hours 2000 2000

Which machine will give the lower cost per unit of production, if run for the whole year? If only 4000pieces are to be produced in a year, which machine would give the lower cost per piece.


103

Solution:

Computation of cost per unit of production of machines

Particulars Machine A Machine B

Interest (fixed cost) Rs. 7,500 Rs. 12,000

Variable cost (hourly charge x Rs. 20,000 Rs. 16,000

annual operating hours)

Total cost Rs. 27,500 Rs. 28,000

Total Output 5 x 2000 = 10,000 8 x 2000 = 16,000

Unit cost Rs. 2.75 Rs. 1.75

If the output is 4000 units per annum

Particulars Machine A Machine B

Interest (fixed cost) Rs. 7,500 Rs. 7,500

Variable cost (10 x 800) Rs. 8,000 Rs. 4,000

(8 x 500)

Total cost Rs. 15,500 Rs. 16,000

Unit cost Rs. 3.89 Rs. 4.00

Note: 800 hours will be required to produce the commodity with machine A and 500 hrs on machine B.

Problem : 13

ABC. Co. has developed a forecast for the group of items that has the following demand pattern

Quarter Demand Cumulative demand

1 270 270

2 220 490

3 470 960

4 670 1630

5 450 2080

6 270 2350

7 200 2550

8 370 3920

The firm estimates that it costs Rs. 150 per unit to increase production rate Rs. 200 per unit to decreasethe production rate, Rs. 50 per unit per quarter to carry the items in inventory and Rs. 100 per unit ifsubcontracted. Compare the costs of the pure strategies.

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Solution:

Different pure strategies are

Plan I In this pure strategy, the actual demand is met by varying the work force size. This means thatduring the period of low demand, the company must fire the workers and during the period of highdemand the company must hire workers. These two steps involve associated costs. In this strategy,the production units will be equal to the demand and values in each period. The cost of the plan iscomputed in the table below,

Quarter Demand Cost of increasing Cost of decreasing Total cost ofProduction level (Rs) Production level (Rs) plan (Rs.)

1 270 — — —

2 220 — 50 x 200 = 10,000 10,000

3 470 250 x 150 = 37,500 — 37,500

4 670 200 x 150 = 30,000 — 30,000

5 450 — 220 x 200 = 44,000 44,000

6 270 — 180 x 200 = 36,000 36,000

7 200 — 70 x 200 = 14,000 14,000

8 370 170 x 150 = 25,500 — 25,500

Total 1,97,000

Plan II In this plan, the company computes the average demand and sets its production capacity tothis average demand. This results in excess of units in some periods and also shortage of units duringsome other periods. The excess units will be carried as inventory for future use and shortage of unitscan be fulfilled using future inventory. The cost of the plan II is computer in the table. The plan incursa maximum shortage of 255 units during 5 periods. The firm might decide to carry 255 units from thebeginning of period 1 to avoid shortage. The total cost of the plan is Rs. 96,000.

Quar- Demand Cumu Production Cumu. Inventory Adjusted Cost ofter forecast lative level prod. inventory with holding

demand level 255 at inventorybeginning Rsof period 1

1 270 270 365 365 95 350 17,5002 220 490 365 730 240 495 24,7503 470 960 365 1095 135 390 19,5004 670 1630 365 1460 –170 85 4,2505 450 2080 365 1825 –255 0 06 270 2350 365 2190 –160 95 4,7507 200 2550 365 2555 5 260 13,0008 370 3920 365 2920 0 255 12,750

Total 96,500


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Plan III

The additional demand other than the normal capacity is met by subcontracting. The cost of the planIII amounts to Rs. 1,32,000 as shown in table below.

Quarter Demand Production Subcontract Incremental cost @forecast units units Rs. 100/units

1 270 200 70 70 x 100 = 7,000

2 220 200 20 20 x 100 = 2,000

3 470 200 270 270 x 100 = 27,000

4 670 200 470 470 x 100 = 47,000

5 450 200 250 250 x 100 = 25,000

6 270 200 70 70 x 100 = 7,000

7 200 200 0 0

8 370 200 170 170 x 100 = 17,000

Total = 1,32,000

The total cost of pure strategies is given below. On observation Plan II (Changing inventory levels)has the least cost.

Plan Total cost (Rs)

Plan I 1,97,000

Plan II 96,500

Plan III 1,32,000

Problem: 14

A company manufactures the consumer durable products and the company intends to develop an aggregateplan for six months starting from January through June. The following information is available.

Demand and working days.

Month Jan Feb Mar Apr May June

Demand 500 600 650 800 900 800

Working day 22 19 21 21 22 20

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Cost details

Materials Rs. 100/ unit

Inventory carrying cost Rs. 10/unit/month

Cost of stockout Rs. 20/unit/month

Cost subcontracting Rs. 200/unit

Hiring and training cost Rs. 50/worker

Lay off cost Rs. 100/worker

Labour hours required Rs. 4/unit

Regular time cost (forfirst 8 hrs) Rs. 12.50/hours

Overtime cost Rs. 18.75/hr

Beginning inventory 200 units

Safety stock required Nil

Work out the cost of the following strategies

1. Produce exactly to meet demand — vary the work force.

2. Constant work force — vary inventory and allow shortages

3. Constant work force and use subcontracting.

Solution:

Strategy I: Produce exactly to meet demand by varying work force.

Assumption: Opening workforce equals the first month’s requirements.

Table: Aggregate production planning requirements

Jan Feb Mar Apr May June Total

Beginning Inventory 200 0 0 0 0 0

Forcasted demand 500 600 650 800 900 800

Production requirement 300 600 650 800 900 800(demand + safety stock –beginning inventory

Ending inventory 0 0 0 0 0 0(beginning inventory+Production requirement –Demand forecast)


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Plan I — Exact Production, vary Work Force


Production requirement 300 460 650 800 900 800

Production hours required 1200 2400 2600 3200 3600 3200(production requirement× 4 hr/unit)

Working days per month 22 19 21 21 22 20

Hours per month per worker 176 152 168 168 176 160(working days × 8 hrs/day)

No. of workers’ required 7 16 15 19 20 20(production hrs required +hrs per month per worker

New worker hired (assuming 0 9 0 4 1 0opening work force equal to firstmonths requirement of 7 workers)

Hiring cost 0 450 0 200 50 0 700(workers hired × Rs. 50)

Workers laid off 0 0 1 0 0 0

Lay off cost 0 0 100 0 0 0 100(workers laid off × 100)

Regular production cost(production hrs required 15,000 30,000 32,500 40,000 45,000 40,000 2,02,500× 12.50 Rs./hrs)

Total 2,03,300

Plan II — Constant Work Force, vary Inventory and Stockout

* Assume a constant work force of 10.


Beginning inventory 200 140 – 80 – 310 – 690 –1150


Production hrs available 1760 1520 1680 1680 1760 1600(working days/month×8 hrs/day× 10 workers)

Actual production 440 380 420 420 440 400(production hrs available÷ 4 hours/unit)

108



Forecasted demand 500 600 650 800 900 800

Ending inventory (beginning 140 – 80 – 310 – 690 –1150 –1550inventory + actual)

Shortage cost 0 1600 6200 13800 23000 31000 75600(unit short × Rs.20/unit)

Units excess (ending inventory 140 0 0 0 0 0 – safety stock)

Inventory cost (unit excess × 10) 1400 0 0 0 0 0 1400

Regular production cost 22000 19000 21000 21000 22000 20000 125000(production hrs required × 12.50 Rs./ hrs)

Total 202000

Plan III — Constant Work Force Subcontract


Production requirement 300 460 650 800 900 800


Production hrs available 1760 1520 680 1680 1760 1600 (working days × 8 hrs/day× 10 workers)

Actual production (production 440 380 420 420 440 400hrs available ÷ 4 hours per unit)

Unit subcontracted (production 0 220 230 380 460 400requirements – actual production)

Subcontracting cost 0 8000 23000 38000 46000 40000 155000(units subcontracted ×Rs.100)

Regular production cost 22000 19000 21000 21000 22000 20000 125000(production hrs required× 12.50 Rs./hrs)

Total 280000

Note: Assume a constant work force of 10.

600 - 140 = 460 units of beginning inventory in February.


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Summary of the Plans

Plan Hiring Lay off Subcon– RT. Short– excess Totaltract prod age inven. cost

Plan I – Exact production 700 100 – 2,02,500 – – 2,03,300vary work force

Plan II – Constant work – – – 1,25,000 75,600 1,400 2,02,000force vary inventory

and shortagesPlan III – Constant work – – 155000 1,25,000 – – 2,80,000

Problem: 15

A company is considering the expansion of a manufacturing process by adding more 1-Ton capacity furnaces.Each batch (1 ton) must undergo 30 minutes of furnace time, including load and unload operations. Howeverthe furnace is used only 80% of the time due to power restriction in other parts of the system. The requiredoutput for the new layout is to be 16 tons/shift (8 hours). Plant (system) efficiency is estimated at 50% ofsystem capacity.

(a) Determine system capacity and the number of furnaces required

(b) Estimate the percentage of time, the furnaces will be idle.

Solution:

(a) Required system capacity

= 16 tons/shift

0.5 = 32 tons / shift =

32 8 0.8× = 5 tons / hour

Individual furnace capacity = 1 ton

0.50 hour= 2 ton/hour per furnace

Number of furnaces required = 5 tons/hour

2 ton / hour per furnace = 2.5 (say) ≈ 3 furnaces.

(b) Percentage of idle time:

Total hours available / shift = 3 furnaces × 8 hours = 24 furnace-hour

Total hours of actual use/shift = (24 -8) = 16 ton × 0.5 hour/ton = 8 furnace-hour

Idle time = 16 furnace-hour % of idle time = 16/24 = 67%

Problem: 16

Annual demand for a manufacturing company is expected to be as follows

Units demanded 8,000 10,000 15,000 20,000

Probability 0.50 0.20 0.20 0.10

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Selling price is Rs. 35 per unit. The existing manufacturing facility has annual fixed operating cost ofRs. 2,06,000. Variable manufacturing costs are Rs. 7.75 per unit at the 8000 unit output level, Rs. 5 atthe 10,000 unit level, Rs. 5.33 at the 15000 unit level and Rs: 7.42 at the 20,000 unit output level.

An expanded facility under consideration would require Rs. 2,50,000 fixed operating costs annually.Variable costs would average Rs. 9.40 at the 8000 unit level, Rs. 5.20 at 10,000 unit level, Rs. 3.80 at the15000 unit level, and Rs. 4.90 at the 20000 level.

To maximise net earnings, which size facility should be selected?

Solution:

Expected net revenue of existing facility:

Expected variable cost

= [7.75×8000×0.5 + 5×10000×0.2 + 5.33×15000×0.2+ 7.42×20000×0.1]

= 31000+10000+15990+14840=Rs.71.830

Expected total cost = fixed cost + variable cost = 2,00,000 + 71,830 = Rs. 2,71,830

Expected sales = 35 [8000×0.5+10000×0.2 + 15000×0.2 + 20000×0.1]= Rs.3,85,000

Expected net revenue = 3,85,000-2,71,830 = Rs. 1,13,170

Expected net revenue of expanded facility: Expected variable cost

= [9.40×8000×0.5 + 5.20×10,000×0.2 + 3.80×15000×0.2 + 4.90×2000×0.1 ]

= 37600+10400+11400+980=Rs. 60,380

Expected total cost = fixed cost + variable cost = 2,50,000 + 60,380 = Rs.3,10,380 Expected net revenue= 3,85,000-3,10,380 = Rs. 74,620. Therefore, the existing facility maximises expected net earnings.

Problem: 17

A manufacturer has the following information on its major product

Regular - time production capacity = 2600 units/period.

Over time production costs = Rs. 12 per unit.

Inventory costs = Rs. 2 per unit per period (based on closing inventory)

Backlog costs = Rs. 5 per unit per period.

Opening inventory 400 units.

Demand (in units) for periods 1, 2, 3, 4, is 4000, 3200, 2000 and 2800 respectively. Develop a level outputplan that yields zero inventory at the end of period 4. What costs result from this plan?


111

Solution:

Period Demand Output Closing Inventory Regular output Overtime(units) (units) (units) (units) output

400

1 4000 2900 –700 2600 300

2 3200 2900 –1000 2600 300

3 2000 2900 –100 2600 300

4 2800 2900 0 2600 300

Average 3000 Total(–)1800

Total Cost = Overtime + inventory + backlogs = (300×4×12) + (0×2) + (1800 ×5) = Rs.23,400.

Problem: 18

M Ltd. produces a product which has a 6-month demand cycle, as shown. Each unit requires 10 workerhours to be produced, at a labour cost of Rs. 6 per hour regular rate (or Rs. 9 per hour overtime). The totalcost per unit is estimated at Rs. 200, but units can be subcontracted at a cost of Rs. 208 per unit. There arecurrently 20 workers employed in the subject department and hiring and training costs for additionalworkers are Rs. 300 per person, whereas layoff costs are Rs. 400 per person. Company policy is to retain asafety stock equal to 20% of the monthly forecast, and each month’ s safety stock becomes the openinginventory for the next month. There are currently 50 units in stock carried at a cost of Rs. 2 per unit-month.Stockouts have been assigned a cost of Rs. 20 per unit-month.

January February March April May June

Forecast demand 300 500 400 100 200 300

Work days 22 19 21 21 22 20

Work hour at 8 per day 176 152 168 168 176 160

Three aggregate plans are proposed.

Plan I. Vary the work-force size to accommodate demand.

Plan II. Maintain a constant work-force of 20, and use overtime and idle times to meet demand.

Plan III. Maintain a constant workforce of 20 and build inventory or incur a stock out cost. The firmmust begin January with the 50-unit inventory on hand.

Compare the costs of the three plans.

Solution:

We must first determine what the production requirements are as adjusted to include a safety stock of20 per cent of next months forecast. Beginning with a January inventory of 50, each subsequent month’sinventory reflects the difference between the forecast demand and the production requirement of theprevious month.

112


January February March April May June

Forecast demand 300 500 400 100 200 300

Work days 22 19 21 21 22 20

Work hour at 8 per day 176 152 168 168 176 160

Plan I (Vary workforce size)

Jan Feb Mar Apr May June Total(in Rs.)

Production Required 310 540 380 40 220 320

Production hrs. Required 3100 5400 3800 400 2200 3200

Available Hrs./worker 176 152 168 168 176 160

No. of worker Required 18 36 23 3 13 20

No. of worker Hired – 18 – – 10 7

Hiring Cost 5400 – 3000 2100 10.500

No. of workers laid-off 2 – 13 20 – –

Lay off cost 800 – 5200 8000 – – 14,000

Plan II (Use overtime and idle time) (based on constant 20-work force)

Jan Feb Mar April May June Total(in Rs.)


Production hrs. Required 3100 5400 3800 400 2200 3200

Available Hrs./worker 176 152 168 168 176 160

Total Available Hrs. 3520 3040 3360 3360 3520 320

O.T. hrs. Required 2360 440 – 0

O.T. Premium – 7080 1320 – – 0 8,400

Idle hours 420 – – 2960 1320 0

Idle Time Cost 2520 – – 17,760 7920 0 28,200


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Plan III (Used inventory and stockouts on a constant 20 - worker force)

Jan Feb Mar April May June Total(in Rs.)


Cumulative 310 850 1230 1270 1490 1810Requirement

Available Hours 3520 3040 3360 3360 3520 3200

Unit produced 352 304 336 336 352 320

Cumulative Production 352 656 992 1328 1680 2000

Units Short — 194 238 — — —

Shortage cost — 3880 4760 — — —

Excess units 42 — — 58 190 190 8,640

Inventory Cost 84 — — 116 380 380 960

Note that Plan III assumes that a stockout cost is incurred if safety stock is not maintained at prescribelevel of 20% of forecast. The firm is in effect managing the safety stock level to yield a specified degreeof production by absorbing the cost of carrying the safety stock as a policy decision.

Summary:

Plan I - 10,500 (Hiring) + 14,000 (Layoff) = Rs.24,500

Plan II - 8,400 (OT) + 28,200 (IT) = Rs. 36,600

Plan III - 8,640 (Stockout) + 960 (Inventory) = Rs. 9,600

Thus, Plan III is the preferred plan.

Problem: 19

X Garment Products produces garment. While planning for next year production following demand(quarterwise) pattern was noticed.

Quarter I II III IV

Demand 7000 10000 9000 10000

At present it is running in single shift operation having rate of production 80 units. As and whenrequired X.G.P. runs a second shift by hiring extra workers, in which the production is only 60 units.The extra workers, once hired, must be kept for any period equal to a quarter or its’ multiples. Thereis also a provision for giving over time to the workers, which is limited to 25% of the regular hours.However, the O.T. provision is only for the quarters where the production is run in a single shift. Theproductivity during O.T. is 20% more than that during regular time. The O.T. wages are quite attractive,being at a premium of 50% over the normal wages, which are Rs. 150 per day. X.G.P pays the samewages to all its workers including the temporary ones hired for the second shift. In each shift 20workers are required.

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The cost of change over from a single shift working to double shift working is Rs.30000, and that froma double shift to a single shift working is only Rs.20000 These change over costs are onetime costs,incurred only at the time the shift working changes are made.

The cost accounting section has worked out the inventory holding cost for the products which comesto Rs.30 per unit per month. A quarter is of three months and every month consists of approximately25 working days.

X.G.P.’s back ordering costs in it are quite heavy, at Rs. 100 per unit per month. An order once delayedcan only be accepted in the next quarter or one next to it, i.e. multiple of a quarter.

There is an initial inventory of 1000 units, which is the amount of safety stock that is required to bekept at all times.

Compute and compare the aggregate production plans which provide for:

1. Production at a continuous rate of 9000 units per quarter.

2. Running a single-shift for the first half of the year and a double shift for the secondhalf of the year. (Wherever possible, use OT to the maximum)

Solution:

(i) Aggregate Plan-I (Constant rate of production)

In single shift possible unit/quarter=80 units/day × 75day/quarter=6000 units /quarter

Which is not sufficient to reach constant 9000 units/quarter

So OT option may be explored

= (25 ÷100 × 75 day/quarter) × (120/100) × 80 units/day = 1800 units/quarter.

By inclusion of OT option it was not possible to reach 9000 unit/quarter.

Introducing 2nd shift operation:

The production per quarter = 60 units /day × 75days/quarter = 4500units/quarter

As per given condition as 2nd shift operation is put in force option of O.T. to be discarded.

Thus total production possible in double-shift = 6000+4500 units/quarter.

The relevant cost of 9000 unit per quarter. Regular time wages = 20 × 2 × 150 × 75 = Rs.4,50,000 Cost ofchange over from single shift to double shift = Rs.30,000/-


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Cost of inventory carrying

Quarter 1 2 3 4

Opening (inventory) 1000 3000 2000 2000

Production–(units) 9000 9000 9000 9000

Demand–(units) 7000 10000 9000 10000

End inventory–(units) 3000 2000 2000 1000

Average inventory (units) 2000 2500 2000 1500

Cost of carrying inventory (Rs) 30×3×2000 30×3×2500 30×3×2000 30×3×1500

=1,80,000 =2,25,000 =1,80,000 = 1,35,000

The total relevant costs over the entire planing

= Regular time wage for year + O.T. + Costs of changeover + inventory cost + Backlog costs

= Rs.4,50,000×4 + 0 + Rs.30,000 + Rs. 1,80,000 + Rs.2,25,000 + Rs. 1,80,000 + Rs. 1,35,000 + 0

= Rs.25,50,000

Aggregate Plan 2 (Single shift for quarter I & II, double shift for quarter III and IV, O.T. to be used tomaximum)

Production Plan

Quarter 1 2 3 4

Regular time production 6000 6000 10500 10500

O.T. 2400 2400 – –

Quarter-I

Production 8400

Opening invetory 1000

9400 units

Demand 7000

Inventory at the end of quarter 2400 units

Average inventory at the end of quarter = (1000 + 2400) ÷2 = 1700 units.

Quarter-II

Production 8400

Opening invetory 2400

10800 units

Demand 10000


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[Here the stock is lower than safety stock] Average inventory

= (2400 + 800) ÷ 2 = 1600 units

Quarter-III

Now the safety stock has to be made up to the 1000 units, i.e. 200 units (1000-800) are demanded over9000 unit demanded.

Production 8400

Beginning invetory 2400

10800 units

Demand 10000


Average inventory = (800 + 1300) ÷ 2 = 1050 units.

Quarter -IV

Production 8400

Initial invetory 2400

10800 units

Demand 10000

Quarter End inventory 1300 units

Average inventory = (1300+1800) ÷ 2 = 1550 units.

The relevant costs for aggregate plan-2 Regular time wages

= 2 single shift + 2 double shift

= 2,25,000×2+4,50,000×2 = Rs. 13,50,000

Overtime wages for 1st 2 quarters = 225000 × (25/100) × (150/100) × 2

= Rs. 168750

Inventory carrying cost = 3 × 30× (1700+ 1600+ 1050+1550)

= Rs. 5,31,000

Change over cost = Rs. 30000

Total relevant cost = Rs.(13,50,000+1,68,750 + 5,31,000 + 30,000)

= Rs. 20,79,750

The aggregate plan-2 is less costly.


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Problem: 20

Machines A and B are both capable of manufacturing a product. They compare as follows:

Machine A Machine B

Investment Rs. 50,000/– Rs. 80,000/–

Interest on capital invested 15% per annum 15% per annum

Hourly charges (Wages + Power) Rs. 10/– Rs. 8/–

No. of pieces produced per hour 5 8

Annual operating hours 2,000 2,000

(i) Which machine will have the lower cost per unit of output, if run for the whole year?

(ii) If only 4000 pieces are to be produced in a year, which machine would have the lower cost perpiece?

(iii) Will your answer to (i) above vary if you informed that 12.5% of the output of machine B getsrejected at the inspection stage. If so, what would be the new solution?

Solution:

Data Machine A Machine B

Annual interest charges Rs. 50,000 × 100

15Rs. 80,000 ×

100

15

= Rs.7,500/– = Rs. 12,000/–

Annual operating charges Rs. 10×2,000 Rs. 8 × 2,000

= Rs. 20,000 = Rs. 16,000

Total annual charge 7,500 + 20,000 12,000 + 16,000

= Rs. 27,500 = Rs. 28,000

Annual production (units) 5 ×2,000 = 10,000 nos 8 ×2,000 = 16,000 nosfor 2000 hours

Cost per unit = 10,000

27,500 = Rs2.75 = 16,000

28,000 = Rs1.75

Machine ‘B’ gives the lower cost per unit if run for the whole year (for 2000 hours),

(ii) If only 4000 pieces are to be produced in an year:

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Data Machine A Machine B

Operating hours required for8

4,000=500hrs.

8

4,000=500hrs.

Operating charges Rs. 10×500= Rs. 5,000/– Rs.8 × 500 =Rs.4,000/–

Interest charges Rs. 7,500/– Rs. 12,000/–

Total annual charges Rs.(5000 + 7500) Rs.(4000 + 12000 )

= Rs. 12,500 = Rs. 16,000

Cost per unit = Rs.3.125/– = Rs. 4/–

Machine ‘A’ gives lower cost per unit.

(iii) If 12.5% of output of Machine B is rejected, net annual production

from Machine B = 16,000 x (100-12.5)

100 = 16,000 x

87.5 100

=14,000

Cost per unit = 28,00014,000 = Rs. 2/-

Even though, unit cost of production on Machine B increases from Rs. 1.75 to Rs. 2.0, still machine Bcontinues to be cheaper, if used for 2000 hours in the year.

Problem:21

Methods P and Q are both capable of manufacturing a product. They compare as follows

Data Method P Method Q

Fixture – cost Rs. 24,000/– Rs. 16,000/–

– life 6 months 4 months

Tooling – cost Rs. 2,500/– Rs. 4,800/–

– life 300 pieces 500 pieces

Processing time per piece 6mts 4 mts

The annual requirement is 1500 nos. Operating cost per hour of the process is Rs. 128 for both processes.Material cost is same in each case.

Which method would you choose for production during a period of one year?

Solution:

Data Method P Method Q

Cost of manufacture per year

Fixture cost Rs. 24,000 x 2 = Rs. 48,000/–

= Rs. 48,000/– Rs. 16,000x3


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(2 nos of fixtures are required per year in method P and 3 nos required in method Q)

Tooling cost = 2,560×300

Rs.1,500= 4,800×

300

Rs.1,500

= 2560 x5 = 4800 x 3

= Rs.12,800 = Rs.14,400

Operating hours to produce 1500 nos.1500 6

60×

=150hrs.1500 4

60×

=100hrs.

Operating cost per year Rs. 128x150 Rs. 128x100

= Rs. 19,200/– = Rs. 12,800/–

Total manufacturing cost per year Rs. 48,000 /– Rs. 48,000 /–

Rs. 12,800 /– Rs. 14,400 /–

Rs. 19,200 /– Rs. 12,800 /–

Rs. 80,000 /– Rs. 75,200 /–

Since method Q is cheaper than method P, method ‘Q’ is the choice for production during the wholeone year/period.

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2.4 Process Planning

A process is a sequence of activities that is intended to achieve some result, typically to create added valuefor the customers.

Types of Processes:

(i) Conversion Processes i.e., converting the raw materials into finished products (for example, convertingiron ore into iron and then to steel). The conversion processes could be metallurgical or chemical ormanufacturing or construction processes.

(ii) Manufacturing Processes can be:

(a) Forming Processes, (b) Machining Processes and (c) Assembly Processes.

(iii) Testing Processes which involve inspection and testing of products (some times considered as part ofthe manufacturing processes).

Forming Processes include foundry processes (to produce castings) and other processes such as forging,stamping, embossing and spinning. These processes change the shape of the raw material (a metal) intothe shape of the work piece without removing or adding material.

Machining Processes comprise metal removal operations such as turning, milling, drilling, grinding,shaping, boring etc.

Assembly Processes involve joining of parts or components to produce assemblies having specific functions.

Process Planning is defined as the systematic determination of methods by which a product is to bemanufactured economically and competitively. It consists of selecting the proper machines, determiningthe sequence of operations, specifying the inspection stages, and tools, jigs and fixtures such that the productcan be manufactured as per the required specification. The detailed process planning is done at eachcomponent level.

After the final design of the product has been approved and released for production, the Production Planningand Control department takes the responsibility of Process Planning and Process Design for convertingthe product design into a tangible product. As the process plans are firmly established, the processing timerequired to carryout the production operations on the equipments and machines selected are estimated.These processing times are compared with the available machine and labour capacities and also againstthe cost of acquiring new machines and equipments required, before a final decision is made to manufacturethe product completely in house or any parts or sub assemblies must be outsourced.

In transformation of raw materials into finished products, several questions need to be answered; such as:

1. What will be the production quantity?

2. What are characteristics of the products to be manufactured?

3. The availability of equipment and what kinds of equipment are to be purchased and what will be theinvestment?

4. What kinds of labour are required?

5. What should be the level of automation?


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6. Make or buy the components required?

Once these questions are answered, the process planning activity can be carried out with minute details asto how each component can be manufactured.

Process Planning is concerned with planning the conversion processes needed to convert the raw materialinto finished products. It consists of two parts – (i) Process design and (ii) Operations design.

The two terms are explained as under:

Process planning establishes the shortest route that is followed from raw material stage till it leaves as afinished part or product.

The activities that are associated with process planning are:

• List of operations to be performed and their sequence.

• Specifications of the machines and equipment required.

• Necessary toolings, jigs and fixtures.

• Gives the manufacturing details with respect to feed, speed, and depth of cut for each operation to beperformed.

• It gives the estimated or processing time of operations.

All the above information is represented in the form of a document called process sheet or route sheet.

The information given in the process sheet can be used for variety of activities.

• It becomes the important document for costing and provides the information on the various detailslike set-up and operation times for each job.

• The machine and manpower requirements can be computed from the set-up and operational times.

• Helps to carryout scheduling.

• The material movement can be traced.

• It helps in cost reduction and cost control.

• It helps to determine the efficiency of a work centre.

Factors affecting process planning

(i) Volume (quantity) of production.

(ii) Delivery dates for components or products.

(iii) Accuracy and process capability of machines.

(iv) The skill and expertise of manpower.

(v) Material specifications.

(vi) Accuracy requirements of components or parts.

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Steps in Process Planning

1. Detailed study of the component drawings to identify the salient features that influence processselection, machine selection, inspection stages and toolings required.

2. List the surfaces to be machined.

3. The surfaces to be machined are combined into basic operations. This step helps in selection of machinesfor operation.

4. Determine the work centre, tools, cutting tools, jigs and fixtures and inspection stages and equipment.

5. Determine the speed, feed and depth of cut for each operation.

6. Estimate the operation time.

7. Find the total time to complete the job taking into account the loading and unloading times, handlingtimes, and other allowances.

8. Represent the details on the process sheet.

Process design is concerned with the overall sequences of operations required to achieve the productspecifications. It specifies the type of work stations to be used, the machines and equipments necessary tocarryout the operations. The sequence of operations is determined by (a) The nature of the product, (b) thematerials used, (c) the quantities to be produced and (d) the existing physical layout of the plant.

Operations design is concerned with the design of the individual manufacturing operation. It examinesthe man-machine relationship in the manufacturing process. Operations design must specify how muchlabour and machine time is required to produce each unit of the product.

Process Design-Framework

The process design is concerned with the following: (i) Characteristics of the product or service offered tothe customers, (ii) Expected volume of output, (iii) Kinds of equipments and machines available in thefirm, (iv) Whether equipments and machines should be of special purpose or general purpose, (v) Cost ofequipments and machines needed, (vi) Kind of labour available, amount of labour available and theirwage rates, (vii) Expenditure to be incurred for manufacturing processes, (viii) Whether the process shouldbe capital-intensive or labour-intensive, (ix) Make or buy decision and (x) Method of handling materialseconomically.

Selection of process

Process selection refers to the way production of goods or services is analysing. It is the basis for decisionsregarding capacity planning, facilities (or plant) layout, equipments and design of work systems. Processselection is necessary when a firm takes up production of new products or services to be offered to thecustomers.

Three primary questions to be addressed before deciding on process selections are:

• How much varieties of products or services will the system need to handle?

• What degree of equipment flexibility will be needed?

• What is the expected volume of output?


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Process decisions:

Major process decisions are:

Process choice :

It refers to choice of a particular process, based upon the nature of product. The operations manager has tochoose from five basic process types – (i) Job shop, (ii) Batch, (iii) Repetitive or assembly line, (iv) Continuousand (v) Project.

Vertical integration :

Vertical integration is the degree to which a firm’s own production system handles the entire supply chainstarting from procurement of raw-materials to distribution of finished goods.

Two directions of vertical integration are:

(a) Backward integration which represents moving upstream toward the sources of raw-materials andparts, for example, a steel mill going for backward integration by owning iron ore and coal mines anda large fleet of transport vehicles to move these raw materials to the steel plant.

(b) Forward integration in which the firm acquires the channel of distribution (such as having its ownwarehouses, and retail outlets).

Procedure for process planning and design

1. The inputs required comprise the product design information, production system information andproduct strategy decisions.

2. Process planning and design starts with selection of the types of processes, determining the sequenceof operation, selection of equipment, tooling, deciding about the type of layout of facilities andestablishing the control system for efficient analysing of resources to achieve most economicalproduction of the product.

3. The outputs are specific process plans, route sheets, flow charts, assembly charts, installation ofequipments, machinery, material handling systems and providing trained, skilled employees tocarryout the production processes to achieve the desired results.

Process analysis and process flow design :

While analysing and designing processes to transform input resources into goods and services, certainquestions need to be asked. They are:

• Is the process designed to achieve competitive advantage in terms of differentiation, response or lowcost?

• Does the process eliminate steps that do not add value?

• Does the process analysing customer value as perceived by the customer?

• Will the process enable the firm to obtain customer orders?

A number of tools help production manager to understand the complexities of process design and redesign.

Some of such tools are: (i) Flow diagram, (ii) Assembly charts, (iii) Process charts and (iv) Operation andRoute sheet.

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These tools are discussed in the following paragraphs.

Flow Diagrams: It is a drawing used to analyse the movement of people or material or product tounderstand, analyse and communicate the process to others.

Assembly Charts: Assembly charts are used to provide an overall macro view of how materials and subassemblies are assembled to form finished products. These charts list all major materials, components, subassembly operations, inspections and assembly operations.

Process Charts: A process chart is understood as a graphic representation of events and information relatingto them during a series of actions or operations.

Operation process charts are similar to assembly charts except that they include specifications for thecomponents as well as operating and inspection times and thereby provide more instruction on how toproduce an item.

The operation analysis and routing sheets or simply the route sheets specify precisely how to produce anitem by identifying the equipment and tools to be used, the operations to be carried out and their sequenceto be followed and the machine set up and run-time estimate.

Purpose of Process Charts

Process charts can present a picture of a given process so clearly that every step of the process can beunderstood by those who study the charts.

Process charts may be effective in process analysis and may help in detecting inefficiencies of the processescurrently adopted.

Types of Process Charts

Process charts can be classified as operation process charts, flow process charts, worker-machine/man-machine charts and activity charts or multiple activity charts.

(a) Operation Process Chart

Operation process chart, the basic process chart is a graphic representation of the points at which materialsare introduced into the process and of the sequences of inspections and all operations except those involvedin material handling. It includes information considered desirable for analysis such as time required andlocation.

Flow Process Chart

Flow process chart is a graphic representation of all operations, transportations, inspections, delays andstorages during a process and includes information for analysis such as time required and distance moved.

It is especially useful in detecting hidden, non-productive costs such as delays, temporary storages anddistances travelled.

Worker-Machine Chart or Man-Machine Chart or Multiple Activity Chart

Worker-machine charts or man-machine charts are graphical representation of simultaneous activities of aworker and the machine or equipment he or she operates. These charts help identify idle time and cost ofboth workers and machines. Alternative worker-machine combinations can be analysed to determine themost efficient arrangement of worker-machine interaction for carrying out a job.


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Worker-machine charts show the time required to complete tasks that constitute a work cycle. A cycle isthe length of time required to progress through one complete combination of work activities.

Operation Analysis and Route Sheet: An operation and route sheet specifies operations and process routingfor a particular part and assembly. It conveys such information as the type of equipment, tooling, andoperations required to complete the part, equipment setup time and operation time etc.

Process improvement

It is a systematic study of the activities and flows of each process to improve the process. Once the processis thoroughly understood, it can be improved.

Process improvement becomes necessary because of relentless pressure to provide better quality at a lowerprice. The basic techniques for analysing the processes such as flow diagrams and process charts are usefulfor understanding the processes and improve them. Improvements can be made in quality, through-puttime, cost, errors, safety and on-time delivery.

Process improvement is necessary when:

(i) the process is slow in responding to the customer,

(ii) the process introduces too many quality problems or errors,

(iii) the process is costly,

(iv) the process is a bottleneck, with work accumulating and waiting to go through it, and

(v) the process involves waste, pollution and little value addition.

Application of BCA in the choice of machines or process

This analysis is the most convenient method for selecting the optimum method of manufacture or machineamongst the competing ones. The cost estimates of the competing methods (both fixed and variable costs)are prepared and a particular quantity N is determined at which the alternatives give the same cost.

If the quantity to be manufactured is less than N the process with lower fixed cost is selectedand if the quantity to be produced is more than N the process with lower variable cost is selected.

Let FA the Annual Fixed Cost of Machine A

FB the Annual Fixed Cost of Machine B

VA – Variable Cost per unit for Machine A

VB = Variable Cost per unit for Machine B

N = Quantity at which costs on both machines will be equal.

∴ Total cost on machine A = Total cost on Machine B for Quantity N

i.e., FA + VA.N = FB+VB.N

or, N(VA-VB) = FB- FA

or, N = B A

A B

F FV V

−−

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The alternative with lower fixed cost will be more economical for manufacturing up to N and once thequantity exceeds N, it is economical to select an alternative with lower variable cost.


Problem 1:

A component can be manufactured either on centre lathe or on a turret lathe. The cost and time informationto process a component is given below.

The tooling costs are to be recovered within a year. There are no repeat orders. The requirements are to bemet in two lots.

(i) Find the quantity at which both alternatives results in equal cost. (BEP)

(ii) Give the decision rule regarding the choice of lathes

(iii) If the quantity required is 800 Nos./year, which of the machine do you propose?

Solution:

Let F1 = Fixed cost for the centre lathe. Fixed cost consists of set-up and tooling up costs.

∴ Fixed cost for centre lathe (F1) = Set-up cost + Tooling up cost

∴ F1 = No. of set-ups/year × set-up time/set-up (hrs) × [(set-up labour rate) + (Depreciation and otherexpense/hr)] + tooling up costs.

= 2 x 3060

x (10 + 2) + 200 = 212 (Rs.)

Similarly, Fixed cost of turret lathe (F2) = 2 90

60×

x (20 + 2) + 500= Rs. 566

Variable cost for centre lathe = V1

V1 = Processing time × [(Labour cost/hr + Depreciation and other cost/hr)]

=1060

(10 + 2) = Rs. 2 / piece

Variable cost for turret (V2)

V2 = 5

60 (20 + 2) = 1.83 Rs. /piece.

Let, Quantity at which both alternatives gives equal cost be ‘N’.

Particulars Centre lathe Turret lathe

� Setup time 30 minutes 120 minutes

� Processing time 10 minutes 5 minutes

� Tooling up cost (Rs.) 200 500

� Labour cost/hr Rs. 2 Rs. 2

� Depreciation and other cost per hour Rs. 10 Rs. 20


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N = 2 1

1 2

F – FV – V =

566 – 2122 – 1.83

= 3540.17

= 2082.3 = 2083.

Thus the break-even quantity is 2083 pieces.

2. Decision Rule

(i) If quantity is below 2083, Centre lathe is preferred because of lower fixed cost.

(ii) For quantities above 2083, turret lathe is preferred.

(iii) For quantity 2083, both are equally feasible select either of the two machines.

Problem 2:

A Company’s fixed and variable costs for manufacturing a component on three alternative machines aregiven below formulate the decision rules for selecting the machines.

Solution:

The total cost = fixed cost + [variable cost / unit × no. of units]

Let, Tc1 , Tc2 and Tc3 be the total costs for engine lathe, capstan lathe and automat respectively

Let, ‘x’ be the number of units to be manufactured Then,

Total cost on Engine lathe (Tc1) = 5 + 0.2x

Total cost on Capstan lathe (Tc2) = 30 + 0.lx

Total cost on Automat (Tc3) = 70 + 0.05x

Now, comparing the Tc1 and Tc2.

∴ 5 + 0.2x = 30 + 0.lx

∴ 0.lx = 25

x = 250

Comparing the Tc2 and Tc3

At B.E.P. Tc2 = Tc3

∴ 30 + 0.1x = 70+0.05x∴ 0.05x = 40∴ x = 800Decision Rules

(i) If the quantity is below 300, select engine lathe

(ii) Between 300 to 800, select capstan lathe

(iii) Above 800, Automat is to be selected.

Fixed Cost (Rs.) Variable Cost (Rs/unit)Engine lathe 5 0.20Capstan lathe 30 0.10Automat 70 0.05

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Problem 3 :

A Job is performed on the milling machine. The following details are given below:

Standard time for job = 6 minutes

No. of jobs to be produced = 70,000 jobs

Machine capacity = 2000 hrs/month

Machine utilisation = 90%

Compute the number of machines required.

Solution:

Standard Time (ST) =660

= 1

10 hrs.

Maximum Production (MP) = 70,000

Machine Capacity (MC) = 2000 hrs/month

Utilisation of capacity (UC) = 0.9

∴ No. of Machines required (N)

N = ST MPMC UC

×× =

0.1 70,0002,000 0.9

×× = 3.88 Machines = 4 Machines

Problem 4:

A company wants to expand the solid propellant manufacturing plant by the addition of more 1 tonnecapacity curing furnace. Each tonne of propellant must undergo 30 minutes of furnace time includingloading and unloading operations. Furnace is used only 80 per cent of the time due to power restrictions.The required output for the new layout is to be 16 tonnes per shift (8 hours). Plant efficiency (system) isestimated at 50 per cent of system capacity.

(a) Determine the number of furnaces required

(b) Estimate the percentage of time the furnace will be idle

Solution

Required system capacity =Actual output

System Efficiency

=16tonnes/shift

0.5= 32.0 tonnes/shift

Reared system capacity (hrs) = ×32tonnes / shift0.8 8 hrs / shift = 5 tonnes/hrs.

Individual furnace capacity =1tonne0.5hr

= 2 tonnes / hr per furnace

≈


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(i) Number of furnaces required (N) =Re quiredfurnance capacityIndividual furnace capacity

N =5 tonnes / hr

2 tonnes / hr / per furnace = 2.5 furnaces (say 3)

(ii) Total Hours available per shift = 3 furnace @ 8 hours = 24 furnaces hrs

(iii) Total Hours of actual use per shift =16 tonnes × 0.5 hr/tonne = 8 furnace hr

Idle hours = 24 - 8 = 16 hours

Percentage idle time = 5hrs idle2hrs total = 66.66% ≈ 67% Idle time.

Problem 5:

A lathe machine is used for turning operation and it takes 30 minutes to process the component. Efficiencyof the lathe is 90 per cent and scrap is 20 per cent. The desired output is 600 pieces per week. Consider 48hours per week. Determine the number of lathes required?

Solution

Assuming 50 weeks in a year.

The output per annum = 600 × 50 = 30,000 units.

The scarp rate is 20%.

∴ The quantity to be produced (including scarp)

Re quiredoutput

(1 Scraprate)− = 30,000(1 0.2)− = 37,500 units

Total time required for turning

= 37,500 × 3060

= 18,750 hours

Production time required with 90 per cent efficiency

= 18,750

0.9 =20833.3 hours

Time available per lathe per annum = 48 × 50 = 2400 hrs

∴ Number of lathes required = = = ≈Timerequired(hrs) 20833.3

8.68 9Timeavailable(hrs) 2400

∴ No. of lathes required = 9

≈

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Problem 6:

An article is processed on three machines A, B and C as shown below:

A

B

C

B

Machine Machine operation time Preparation Cleaningtime (min/day) (min/day)

Time Processing TotalA 2 2.5 4.5 15 10B 3 10 13 30 10C 2 5 5 35 10

A study revealed that if the jigs for machines B and C were to be redesigned, loading and unloadingtimes could be reduced to 2 minutes and 1 minute respectively.

(a) Find the number of pieces produced per day (single shift of 8 hrs).

(b) Costing has shown that unless production is increased by 20 per cent the installation of new jigswould not be worthwhile. Would you recommend redesign of jigs.

(c) If the number to be produced is large, suggest changes in present arrangement and estimate newproduction rate.

Solution:

Machines Processing time (minutes) Preparation and cleaning (min/day)A 2 + 2.5 = 4.5 25B 3 + 10 = 13 40C 2 + 5 = 7 35

(a) Cycle time for the job is 13 minutes.Total production-time available/day = 480 - preparation and cleaning time/dayFor machine B (critical operation) = 480 – 40 = 440 min= 440/13Output/day = 34 pieces

(b) Redesigning the new jigs: The redesigning of new jigs will change the cycle time, i.e., the cycletime is reduced to 12 minutes (from 13 min as the unloading and loading time is reduced by 1minute)

Output/day = 440/12 = 36.6 = 37 pieces

Percentage increase in output = 37 34

34−

× 100 % = 9%

So the redesign of jigs is not justified.

(c) Suggestion if the number of pieces is large: The critical operation (bottleneck) is processing onmachine B which requires 10 minutes. If we introduce one more machine of kind B, then thecycle time will be reduced to 8 min.


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The new output = 4408

= 55 pieces.

Problem 7:

A component is to be processed on two machines lathe and milling machine. The sequence of operation isfirst turning and then milling.

The machine times are given below:

Turning 12 minutes.

Milling 20 minutes.

1. Estimate the number of machines required to machine 2500 components per week if available machinehours per week are 48.

2. What are the steps that you propose to reduce number of machines.

Solution:

1. No. of components required to be machined = 2500/week

The available time (in hr.) per week = 48

Assuming 50 weeks in a year, no. of components to be machined/annum = 2500 × 50 = 1,25,000annum.

The time required to machine 1,25,000 components = 1,25,000 × 1260

= 25,000 hrs

Time available per annum = 48 × 50 = 2400 hrs

No. of lathes required = 10.41 ≈ 11

No. of milling machines required = ×

= ≈

2012,000

60 17.36 182400

2. Steps to reduce the number of machines. Increase the number of shifts. Addition of second shiftis going to reduce the number of machines by 50 per cent.

Problem 8:

The component can be processed on either of the two machines — Turrette lathe and Center lathe. Thetime and cost details are given below.

Computer the breakeven quantity and state the decision rules.

Particulars Turrette Lathe Centre Lathe1. Setup time (hrs) 4.5 0.22. Operation time/piece (min) 4.0 353. Setup cost/hour (Rs.) 350 154. Machining cost/hour (Rs.) 45 25

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Solution:

Breakeven quantity refers to the quantity at which both the alternatives are equally feasible and thetotal cost of both alternatives are equal

Let ‘Q’ be the breakeven quantity

Total cost of production on Turrette lathe = Total cost of production on center lathe

= [Setup cost + operation cost] Turret = [Setup cost + operation cost]Center

= [4.5×350] +4

4560⎡ ⎤× ×⎢ ⎥⎣ ⎦

Q = [0.2 × 15] + 35

2560

⎡ ⎤× ×⎢ ⎥⎣ ⎦Q

or, 1575 + 3Q = 3 + 14.58Q

or, 11.58Q = 1572

or, Q = 136 units.

If the quantity is < 136, Center lathe is preferred.

If the quantity is > 136, Turrette Lathe is preferred.

Problem 9:

A customer is processed through each of the three operations A, B and C, in sequence. The process isdesigned to handle 100 customers a day. The average rate at which each operation can process customersis shown in the figure below:

A B C

(a) For an 8 hour day identify any bottlenecks in the process.

(b) What effect will bottlenecks have on overall output and other operations?

(c) If all the processes are operated for 10 hours per day, is there still a bottleneck?

(d) How will random arrivals affect the output and processing rate of this process?

Solution:

(a) Identification of bottleneck for an 8 hour a day operation.

Operation A, Output for 8 hrs/day = 15 × 8 = 120 customers

Operation B, Output for 8 hrs/day = 10 × 8 = 80 customers

Operation C, Output for 8 hrs/day = 12 × 8 = 96 customers

Looking into the above calculations, the operations B which process 10 customer/hr and 80customers a day becomes the bottleneck.

(b) Effect of the bottleneck on the overall output.

The bottleneck operation decides the capacity and output. Here the capacity of the unit is 80customer/day (due to bottleneck B) even though the process A and C have the output of 120 and96 respectively.

15 Customer /hr 10 Customer /hr 12 Customer /hr


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(i) Effect of the bottleneck on other operation. Bottleneck operation B, makes the waiting timehigher for C. Because of difference in out (C > B), facility has to wait fill the customer comesfrom B.

(ii) In case A, the operation will be completed and because of difference in processing rates,operation A will be idle for 25% time unless it is used for other purpose and the full capacityof A if used creates an inventory between A and B.

(c) If all the processes are operated for 10 hrs, then the designed output of 100 customers/hr. areachieved still the process line is unbalanced as the processing capacities are unequal.

(d) Random arrivals affect the output and processing of customers.

The random arrivals affect the capacity utilisation also both capacity idle time and shortage.

2.5 Project Planning

Planning begins with well-defined objectives. The project team may be drawn from several organizationaldepartments, e.g., engineering, production, marketing, and accounting. Project definition involvesidentifying the controllable and uncontrollable variables involved, and establishing project boundaries.Performance criteria should relate to the project objectives, which are often evaluated in terms of time,cost, and resource utilisation

Project planning is part of project management, which relates to the use of schedules such as Gantt chartsto plan and subsequently report progress within the project environment. Project management is thediscipline of organizing and managing resources (e.g. people) in such a way that the project is completedwithin defined scope, quality, time and cost constraints. A project is a temporary and one-time endeav orundertaken to create a unique product or service, which brings about beneficial change or added value.This property of being a temporary and one-time undertaking contrasts with processes, or operations,which are permanent or semi-permanent ongoing functional work to create the same product or serviceover and over again. The management of these two systems is often very different and requires varyingtechnical skills and philosophy, hence requiring the development of project managements.

The first challenge of project management is to make sure that a project is delivered within definedconstraints. The second, more ambitious challenge is the optimized allocation and integration of inputsneeded to meet predefined objectives. A project is a carefully defined set of activities that use resources(money, people, materials, energy, space, provisions, communication, etc.) to meet the predefined objectives.

Initially, the project scope is defined and the appropriate methods for completing the project are determined.Following this step, the durations for the various tasks necessary to complete the work are listed andgrouped into a work breakdown structure. The logical dependencies between tasks are defined using anactivity network diagram that enables identification of the critical path. Float or slack time in the schedulecan be calculated using project management software. Then the necessary resources can be estimated andcosts for each activity can be allocated to each resource, giving the total project cost. At this stage, theproject plan may be optimized to achieve the appropriate balance between resource usage and projectduration to comply with the project objectives. Once established and agreed, the plan becomes what isknown as the baseline. Progress will be measured against the baseline throughout the life of the project.Analysing progress compared to the baseline is known as earned value management.

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Gantt Chart: Gantt Chart is a principal tool used in scheduling and also in some methods of loading. Thischart was originated by the American engineer Henry L. Gantt and consists of a simple rectangular grid,divided by series of parallel horizontal and verticular lines. The vertical lines always divide the horizontalscale units of time. The time units can be in years, months, weeks, days, hours, minutes or even secondsaccording to the work for which it is prepared. In this chart, the time which an activity takes in completingthe task is represented by the horizontal line. The length of the line is drawn in proportion to the durationof time. Generally, the time in the chart should flow from left to right and activities be listed from top tobottom. The progress of the work may be shown by a bar or a line within the uprights of the activitysymbol and its length should represent the amount of work completed. Horizontal lines divide the chartinto sections which can represent various work tasks (work schedule) or work centres (load schedule).When it shows only work tasks-products, orders, or operations to be completed, it is known as WorkSchedule. When it shows the same task opposite the work centres at which they are produced-factories,departments, workshops, machine tools or men it is known as Load Chart.

The units scheduled or loaded on these charts are always the same because these work tasks are known ashaving a known standard time. The work tasks can be represented on the chart by numbers or symbols.The symbols used on the chart may van from company to company.

Network Analysis: Routing is the first step in production planning. In small projects, routing is verysimple. Sequence of operations is almost decided and the operations can be performed one after the otherin a given sequence. But in large project, this is rather a difficult problem. There may be more than oneroute to complete a job. The function of production manager is to find out the path which takes the leasttime in completing the project.

In a big project, many activities are performed simultaneously. There are many activities which can bestarted only at the completion of other activities. In such cases, a thorough study is required to collect thecomplete details about the project and then to find out a new, better and quicker way to get the work donein a decent way. In such cases, the first step is to draw some suitable diagram showing various activitiesand their positions in the project. It should also explain the time to be taken in completing the route fromone operation to the other. It also defines the way in which the delay in any activity can affect the entireproject in terms of both money and time. Such a diagram is called network diagram. A network is a pictureof a project, a map of requirements tracing the work from a departure points to the final completion objective.It can be a collection of all the minute details involved or only a gross outline of general functions.

Important characteristics in a Network Analysis: The following are some important points to rememberin a network analysis:

(i) The objective is to be finished within the specified time otherwise there is a penalty.

(ii) Various activities are to be completed in an order; however, a number of activities are performedsimultaneously while there are many other activities, which can be started only when some otheractivities are completed.

(iii) The cost of any activity is proportional to its time of completion.

(iv) There can be hurdles in the process and the resources to be allocated may be limited. A networkgraph consists of a number of points or nodes, each of which is connected to one or more of the othernodes by routes or edges. It is a set of operations and activities describing the time orientation of acomposite project.


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Procedure for drawing a network diagram: The procedure for drawing a network diagram may beexplained below.

There are three basic questions and the network depends on them.

These questions are:

� Which operation must be completed before each given operation can be started?

� Which activities can be carried out in parallel?

� Which operation immediately succeeds other given activities?

The common practice is simply to work backward through the list of operations, generating the immediatepredecessors for each operation.

Slack and float:

Slack – Slack signifies the freedom for rescheduling or to start the job. It can be calculated by the differencebetween EFT and LFT for any job. A job for which the slack time is zero is known as critical job. The criticalpath can be located by all those activities or events for which slack time is either zero or float time is theleast. The abbreviations EFT and LFT given in the above line have the following explanation.

EFT (Earliest Finish Time) – this is the sum of the earliest start time plus the time of duration for anyevent.

LFT (Latest Finish Time) – It is calculated from the LFT of the head event. For its calculation total projecttime is required. The total project time is the shortest possible time required in completing the project.

Floats – Floats in the network analysis represent the difference between the maximum time available tofinish the activity and the time required to complete it. There are so many activities where the maximumtime available to finish the activity is more than the total time required to complete it. This difference isknown as floats.

Floats may be total, free, and independent:

Total Float: Total float is the maximum amount by which duration time of an activity can be increasedwithout increasing the total duration time of the project. Total float can be calculated as follows:

(i) First, the difference between Earliest Start Time (EST) of tail event and Latest Finish Time (LFT) ofhead event for the activity shall be calculated.

(ii) Then, subtract the duration time of the activity from the value obtained in (i) above to get the requiredfloat for the activity.

The total float can be helpful in drawing the following conclusions:

(a) If total float value is negative, it denotes that the resources for completing the activity are not adequateand the activity, therefore, cannot finish in time. So, extra resources or say critical path needs crashingin order to reduce the negative float.

(b) If the total float value is zero, it means the resources are just sufficient to complete the activity withoutany delay.

(c) If the total float value is positive, it points out that total resources are in excess of the amount requiredor the resources should be reallocated to avoid the delay otherwise the activity will be delayed by somuch time.

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Free Float: It is that fraction from total float of an activity which can be used for rescheduling the activitywithout affecting the succeeding activity. If both tail and head events are given their earliest times, i.e., ESTand EFT the Free Float can be calculated by deducting head slack from total float, i.e.,

Free Float = Total float - Slack time of the head event.

Independent Float: It is the time by which an activity can be rescheduled without affecting the otheractivities – preceding or succeeding. It may be calculated as follows:

= Earliest Start Time (EST) - Earliest Finish Time (EFT) of the activity

or, Independent Float = Free Float -Slack Time of tail event. The basic difference between slack and floattime is that a slack is used with reference to events whereas float is used with reference to activity.

Use of Float Information in Decision Making: The float information can be used in decision-making inthe following ways:

(i) Total float can affect both the previous and the subsequent activities.

(ii) Total float can be used without affecting the subsequent activities.

(iii) Independent float can be used in allocating the resources elsewhere and increasing the time ofsome non-critical activities.

(iv) Negative float signifies reduction in target time to finish the work in time.

Critical Path Method (CPM): The critical path analysis is an important tool in production planning andscheduling.Gnatt charts are also one of the tools of scheduling but they have one disadvantage for whichthey are found to be unsuitable. The problem with Gnatt Chart is that the sequence of operations of aproject or the earliest possible date for the completion of the project as a whole cannot be ascertained. Thisproblem is overcome by this method of Critical Path Analysis.

CPM is used for scheduling special projects where the relationship between the different parts ofprojects is more complicated than that of a simple chain of task to be completed one after the other.This method (CPM) can be used at one extreme for the very simple job and at other extreme for themost complicated tasks.

A CPM is a route between two or more operations which minimises (or maximises) some measures ofperformance. This can also be defined as the sequence of activities which will require greatest normal timeto accomplish. It means that the sequence of activities which require longest duration are singled out. It iscalled at critical path because any delay in performing the activities on this path may cause delay in thewhole project. So, such critical activities should be taken up first.

One of the purposes of critical path analysis is to find the sequence of activities with the largest sum ofduration times, and thus find the minimum time necessary to complete the project. The critical series ofactivities is known as the ‘Critical Path’.

Under CPM, the project is analysed into different operations or activities and their relationship aredetermined and shown on the network diagram. So, first of all a network diagram is drawn. After this therequired time or some other measure of performance is posted above and to the left of each operationcircle. These times are then combined to develop a schedule which minimises or maximises the measure ofperformance for each operation. Thus CPM marks critical activities in a project and concentrates on them.It is based on the assumption that the expected time is actually the time taken to complete the object.


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Thus CPM technique is a very useful analysis in production planning of a very large project.

PERT (Programme Evaluation and Review Technique):

There are so many modem techniques mat have developed recently for the planning and control of largeprojects in various industries especially in defence, chemical and construction industries. Perhaps, thePERT is the best known of such techniques.

PERT is a time-event network analysis technique designed to watch how the parts of a programme fittogether during the passage of time and events. This technique was developed by the special project officeof the U.S. Navy in 1958. It involves the application of network theory to scheduling, problems. In PERTwe assume that the expected time of any operation can never be determined exactly.

Major Features of PERT or Procedure or Requirement for PERT:

The following are the main features of PERT:

(i) All individual tasks should be shown in a network. Events are shown by circles. Each circle representsan event—an event —a subsidiary plan whose completion can be measured at a given time.

(ii) Each arrow represents an activity —the time - consuming elements of a programme, the effort thatmust be made between events.

(iii) Activity time is the elapsed time required to accomplish an event. In the original PERT, three-timevalues are used as follows:

• t1 (Optimistic time): It is the best estimate of time if every tiling goes exceptionally well.

• t2 (Most likely time): It is an estimated time what the project engineer believes necessary to do thejob or it is the time which most often is required if the activity is repeated a number of times.

• t3 (Pessimistic time): It is also an activity under adverse conditions. It is the longest time andrather is more difficult to ascertain.

The experiences have shown that the best estimator of time out of several estimates made by the projectengineer is:

t = 1 2 346

+ +t t t and the variance of t is given by: V (t) =

2

3 1

6−⎛ ⎞

⎜ ⎟⎝ ⎠

t t

(iv) The next step is to compute the critical path and the slack time.

A critical path or critical sequence of activities is one which takes the longest time to accomplish thework and the least slack time.

Difference in PERT and CPM - Although these techniques (PERT and CPM) use the same principles andare based on network analysis yet they are in the following respects from each other:

(i) PERT is appropriate where time estimates are uncertain in the duration of activities as measured byoptimistic time, most likely time, and pessimistic time, whereas CPM (Critical Path Method) is goodwhen time estimates are found with certainty. CPM assumes that the duration of every activity isconstant and therefore every activity is critical or not.

(ii) PERT is concerned with events which are the beginning or ending points of operation while CPM isconcerned with activities.

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(iii) PERT is suitable for non-repetitive projects while CPM is designed for repetitive projects.

(iv) PERT can be analysed statistically whereas CPM not.

(v) PERT is not concerned with the relationship between time and cost, whereas CPM establishes arelationship between time and cost and cost is proportionate to time.

Problem and Solution

Problem: 1

A small project is composed of time activities whose time estimates are given below:

Activities A, B and C can start simultaneously. Activity D follows activity A while E follows B.

Activity D and E are followed by activity G while F is dependent on C, H depends on D and E, while I

depends on F and G.

(i) Construct the network.

(ii) Find the expected duration and variance of each activity.

(iii) Calculate the slack for each event.

(iv) What is the critical path and expected project duration of the project?

(v) The project due date is 28 days, what is the probability of meeting the due date?

(vi) What should be the project duration for the probability of completion of 95%?

Solution:

Expected time Te = 4

6+ +a b m

= and Variance V = 2

6−⎛ ⎞

⎜ ⎟⎝ ⎠

b a

Activity Expected time (Te) Variance (v)

A 3 1

B 5 1

C 5 1

D 2 0

E 6 4

F 7 4

G 5 1

H 8 1

I 7 2

Activity A B C D E F G H I

Optimisctic time 2 2 4 2 2 3 2 5 3

Most likely time 2 5 4 2 5 6 5 8 6

Pessimistic time 8 8 10 2 14 15 8 11 15


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Z = s e–SD

T T=

28 – 230.833

SD=

... Probability of completion = 96.16% (approx)

With 95% probability Z=1.65 = s – 23SD

T

... Schedule time = 33 days

2.6 Progressing and Follow-up

Expediting or progressing ensures that, the work is carried out as per the plan and delivery schedules are met.

Progressing including activities such as status reporting attending to bottlenecks or hold-ups in productionand removing the same, controlling variations or deviations from planned performance levels, followingup and monitoring progress of work through all stages of production, co-ordinating with purchase, stores,tool room and maintenance departments and modifying the production plans and re-plan if necessary.Need for expediting may arise due to the following reasons.a. Delay in supply of materials.b. Excessive absenteeism.c. Changes in design specifications.d. Change in delivery schedules initiated by customers.e. Break down of machines or tools and fixtures.f. Errors in design and process plans.

2.7 Dispatching

Dispatching may be defined as setting production activities in motion through the release of orders workorder, shop order and instructions in accordance with the previously planned time schedules and routings.

Construction of NetworkTE = 3 TE = 5 TE =19TL = 9 TL = 11 TL = 23

Critical path is�

1 �

3 �

5 �

9 �

10

B �

E �

G �

I

Project duration 23 daysStandard deviation σ = ijV = 2.83

B

2

1

2

3

4

5

6

7 8

109

TE = 0 TL = 0

TE = 5 TL = 5

TE = 11 TL = 11

TE = 16TL = 16

TE = 23 TL = 23

TE = 5 TL = 9

TE = 12TL = 16

A

C

D

E

F

G

H

DI

D2

6

7

7

I

5

5

8 3

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Fig. Dispatching

Dispatching also provides a means for comparing actual progress with planned production progress.Dispatching functions include:

a. Providing for movement of raw materials from stores to the first operation and from one operationto the next operation till all the operations are carried out.

b. Collecting tools and fixtures from tool stores and issuing them to the user department or workers.

c. Issuing job orders authorizing operations in accordance with dates and times as indicated inschedules or machine loading charts.

d. Issue of drawings, specifications, route cards, material requisitions and tool requisitions to theuser department.

e. Obtaining inspection schedules and issuing them to the inspection section.

f. Internal materials handling and movement of materials to the inspection area after completingthe operation, moving the materials to the next operation centre after inspection and movementof completed parts to holding stores.

g. Returning fixtures and tools to stores after use.

2.8 Scheduling Technique and Line Balancing Problem

Scheduling: ‘Scheduling’ is the next important function of production planning and control after ‘Routing’.It determines the starting and the completion timings for each of the operations with a view to engageevery machine and operator of the system for the maximum possible time and; without imposingunnecessary burden over them. Scheduling is the determination of the time that should be inquired toperform each operation and also the time that should be required to perform the entire series as routed.Scheduling involves establishing the amount of work to be done and the time when each element of thework will start or the order of the work.

Scheduling technique is an important technique of determining the starting and the completion timings ofeach operation and that of the total manufacturing process so that the man and machines can be utilised tothe maximum.

Dispatching

Centralised Decentralised

Ex. ForemanEx. CentralDespatchSection

M/c. Shop Worker

Issue of order andInstruction to foreman

Issue of orders andInsuruction fromforeman

(Foreman Implacement in each shop) (No. central Instructions here)


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Scheduling depends upon a number of factors, e.g., routing, the method of production, quantity ofproduction, transportation of raw materials, production capacity, the probable data of delivery specifiedby customers in their orders and the past records.

Relationship between ‘Routing’ and ‘Scheduling’. ‘Routing’ and ‘Scheduling’ are independent and eitherof these activities cannot be undertaken independently. It is very difficult to prepare schedules withoutdetermining the routing of sequence of operations. Routing is the prerequisite of scheduling. Unless routeor sequence of operations, tools, equipment and plants and the persons by when operations are to beperformed, are established, the time taken by each operation, the idle time of men and machine and totaltime for the whole process cannot be ascertained in a convincing manner.

Conversely, scheduling is equally important for routing. It is quite difficult to route an item efficientlythrough a plant without consulting previously-designed schedules. The main aim of routing is to pass theitem through the process of manufacture by a route which is the best and the most economical. And aroute or sequence of operations may be considered best which utilises the men, materials and machines tothe maximum and which consumes the shortest time during the process of production. This information(time schedule of each operation) can be obtained from schedules. So, scheduling is necessary for effectiverouting.

Thus, we can conclude that routing and scheduling are inter-related, inter-connected and inter-dependentactivities of production planning and control.

Relationship between Routing and Scheduling: Both are interconnected as scheduling is difficult withoutrouting and routing is also not effective without scheduling. Routing is a prerequisite for scheduling whiletime to be taken ‘may form the basis of routing and that is fixed by scheduling.

Principles of Scheduling:

The principles of scheduling are:

(a) The principle of optimum task size: Scheduling tends to achieve its maximum efficiency when thetask sizes are small and all tasks are of the same order of magnitude.

(b) The principle of the optimum Production plan: Scheduling tends to achieve its maximum efficiencywhen the work is planned, so that it imposes an equal/even load on all the plant.

(c) The principle of the optimum operation sequence: Scheduling tends to achieve its maximum efficiencywhen the work is planned so that the work centers are normally used in the same sequence.

The first principle has a tendency when applied, not only give good results but also to be self-correcting ifit is ignored. For example, if in a functional batch production machine shop the loads imposed by differentoperations vary greatly in length it is possible that it will be necessary to break many of the long operationsinto one or more small batches, in order to get the other orders completed by due date. In effect, thisprinciple only repeats the known advantage of maintaining a high rate of stock turn over, and of singlephase ordering. The second principle merely states that the obvious fact that there will be less idle timeand waiting time, if all the plant is evenly loaded by the production planners, then if some of the machinesare over loaded perhaps because direct labour cost on them are lower and others are idle for part of thetime due to shortage of work. The third principle says about principle of flow. Some times it is also true ifwe sequence some jobs, which need the same machine set up, at a time, this avoids machine ancillary timeneeded, in case, the jobs of the above type are done at different times. For example, consider drilling a 10mm hole in five different jobs may be done at a time so that the set up time required for five jobsindependently at different times are avoided.

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Forms of Schedules:

Here we shall discuss the presentation of production schedules. Depending on the need and use, theSchedules can be prepared in different forms.

A Production Flow Program:

If a number of components or assemblies have to be manufactured for the final assembly line and thosecomponents are to be made concurrently, the production master flow program is prepared taking intoaccount the sequence of operations and the time of starting and ending each component in order to complywith the required date of completion of the product. The necessary document for this is Operation ProcessChart and the Sequence of Operation.

Scheduling Systems:

Scheduling Systems may be classified into four groups as shown below:

(i) Unit scheduling system: This is used for scheduling when jobs are produced one by and are of differenta type that is for job production.

(ii) Batch scheduling system: When jobs are produced to order, in batches, this is used.

(iii) Mass scheduling system: When large number of items of similar type are produced that is in massproduction, this is used.

Unit Scheduling System:

Here we have two types of scheduling, one is Project scheduling and the other is Job Scheduling.

Project Scheduling: Generally, a project consists of number of activities managed by defferent Apartmentsor individual supervisors. It can also be said as a complex output made up of many interdependent jobs.Examples are: Railway coach building, Shipbuilding etc. The scheduling methods used are:

(i) Project Evaluation and Review Technique (PERT),

(ii) Critical Path Method (CPM),

(iii) Graphical Evaluation and Review Technique (GERT).

We can also use Bar charts, GANTT charts, Milestone chart, but these are less superior to the above.

Job Shop Scheduling: In Job shop scheduling, we come across varieties of jobs to be processed on differenttypes of machines. Separate records are to be maintained for each order. Only after receiving the order, onehas to plan for production of the job. The routing is to be specified only after taking the order. Schedulingis done to see that the available resources are used optimally. The following are some of the methods usedfor scheduling. (i) Arrival pattern of the job, (ii) Processing pattern of the job, (iii) Depending on the type ofmachine used, (iv) Number of workers available in the shop, (v) Order of sequencing.

Arrival pattern of the job: This is done in two ways. Firstly, as and when the order is received, it is processedon the principle First in First Out (FIFO). Otherwise, if the orders are received from single customer at differentpoint of time in a week/month, then the production manager pile up all orders and starts production dependingon the delivery date and convenience (This situation is generally known as static situation).

Processing Pattern of the Job: As the layout of Job shops of Process type and there may be duplication ofcertain machines, the production planner, after receiving the order thinks of the various methods of


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converting the requirement of customer / order into a production plan to suit the available facilities.Depending on the process required, there may be backtracking, which is unavoidable. When facilities arebusily engaged, in process inventory may be a common problem.

Machine varieties available: Facilities available in the production shop will affect the scheduling. Herethe size, capacity, precession and other factors of machines will have their influence on the scheduling.

Number of Men in the production shop: Many a time we see that the number of workers available in thejob shops are very much limited, that is sometimes they are less in number than the machines in the shop(these shops are known as labour limited shop). Depending the availability of labour, the scheduling is tobe done. In case the machines available are limited and have more men (known as machine limited shops),then availability of machine dictates the scheduling.

Sequencing rules for single facility: When we have a single facility, and the orders are in queue, then theyare processed depending on the rules mentioned below:

(a) First in first served or first in first out (FIFS/FIFO): Here the jobs are processed as they come in. Thisis commonly observed queue discipline.

(b) Shortest processing time (SPT): The jobs having shortest processing time are processed first. This isjust to avoid formation of queue. For example, when you go for Xeroxing a document, and otherperson comes for Xeroxing a book, then document is Xeroxed and then the book is taken for Xeroxing.

(c) Minimum due date (MDD): Here jobs are processed in ascending order of their available time beforedelivery date. By doing so, we can keep up the delivery promises. To meet the delivery promises, ifnecessary, overtime, sub contracting etc., may be used.

(d) Last come first served or last in first out (LCFS/LIFO): This generally happens in case of inventorystocking and using. When material piles up, the material at the top i.e., material last arrived is usedfirst.

(e) Static slack for remaining operations (SSRO): Static slack is given by: (Due date - Remainingprocessing time/number of remaining operations). Here jobs are processed in ascending order of theoperations.

(f) Dynamic slack for remaining operations (DSRO): Dynamic slack is given by: (Due date - expectedtime of remaining operations / number of remaining operations). Here the jobs are done in ascendingorder of the ratio dynamic slack.

Basic Scheduling Problems:

The production planner may face certain problems while preparing production plans or Schedules. Someimportant problems are discussed below:

(a) Flow production scheduling for fluctuating demand (known smoothening problem),

(b) Batch production scheduling, when products are manufactured consecutively,

(c) The assignment problem,

(d) Scheduling orders with random arrivals and

(e) Product sequencing.

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Problem: 1

A company has two plants A and B with fixed costs of Rs. 50,000 and Rs. 70,000 respectively. Both theplants are designed to produce up to 10,000 units each. The variable costs of two plants at different ofproduction are as follows:

Production (Units) Plant A (Rs.) Plant B (Rs.)

2,500 36,000 29,000

5,000 45,000 39,000

7,500 77,000 51,000

10,000 1,10,000 1,15,000

Find the most economic loading schedule.

Solution:

The fixed costs are irrelevant and only the incremental costs should be considered. Incremental costswill be as follows:

Plant A (Rs.) Plant B (Rs.)Total incremental costs 36,000 29,000

9,000 10,00032,000 12,00033,000 64,000

Per unit incremental costs 14.40 11.603.60 4.0012.80 4.8013.20 25.60

Ans.First 2500 Units BNext 2500 Units ANext 2500 Units BNext 2500 Units A

Problem: 2A company is setting an assembly line to produce 192 units per eight hour shift. The information regardingwork elements in terms of times and immediate predecessors are given

Work element Time (Sec) Immediate predecessorsA 40 NoneB 80 AC 30 D, E, FD 25 BE 20 BF 15 BG 120 AH 145 GI 130 HJ 115 C, I

Total 720


145

(i) What is the desired cycle time?

(ii) What is the theoretical number of stations?

(iii) Use largest work element time rule to workout a solution on a precedence diagram.

(iv) What are the efficiency and balance delay of the solution obtained?

Solution:

The precedence diagram is represented as shown below:

(a) Cycle time = 8 hours1

r 192 units= = 150 sec/unit, where r = output.

(b) Sum of the work elements is 720 seconds, so minimum number of work stations

= = =∑ t 720second/unit4.8 5

cycle time 150sec/ unit station stations

Assignment of work elements to work stations.

(d) Efficiency = t 720

100 100 96%nCt 5 150

× = × =×

∑

Thus, the balance delay is [100 - 96] = 4 percent only.

Station Elements Work element time Cumulative time Idle time forin sec (Sec) station

S1 A 40 40B 80 120 05D 25 145

S2 G 120 120E 20 140 10

S3 H 145 145S4 I 130 130

F 15 145 05S5 C 30 30

J 115 145 05

146


Problem: 3

The company is engaged in the assembly of a wagon on a conveyor. 500 wagons are required per day.Production time available per day is 420 minutes. The other information is given below regarding assemblysteps and precedence relationships. Find the minimum number of work stations, balance delay and lineefficiency.

Solution:

The element times and precedence relationships

(i) Precedence diagram is constructed as per the given details.

(ii) Determination of cycle time

Ct = ×= = =

Production time /day 420 3 25, 20050.4

Output /day 500 500

(iii) Theoretical number of work stations required

N =Total time 195

3.87Cycle time 50.4

= = ≈ 4 workstations

(iv) Assign the elements to work stations based on the largest element time and the work stations tobe required are 5.

Balance made according to largest number of followers task rule.

Task Time (sec) Task that must PrecedeA 45 –B 11 AC 09 BD 50 –E 15 DF 12 CG 12 CH 12 EI 12 EJ 08 F, G, H, IK 09 –

Total 195 –

sec


147

(iv) Efficiency = = × =× ×

T 195100 77.38%

No.of stations C 5 50.4

(v) Balance delay = 100 - 77.38 = 22.62%

Problem: 4

The processing times (tj) in hrs for the five jobs of a single machine scheduling is given. Find the optimalsequence which will minimise the mean flow time and find the mean flow time.

Determine the sequence which will minimise the weighted mean flow time and also find the mean flowtime

Station Task Task time (sec) Idle time (sec)S1 A 45 5.4S2 D 50 0.4S3 B 11 3.4

E 15C 09F 12

S4 G 12 6.4H 12I 12J 8

S5 K 9 41.4

Job (j) 1 2 3 4 5Processing time (tj) hrs 30 8 10 28 16

Solution:

(a) First arrange the jobs as per the shortest processing time (SPT) sequence.

Job (j) 2 3 5 4 1Processing time (tj) hrs 8 10 16 28 30

Job (j) 2 3 5 4 1

Processing time (tj) hrs 8 10 16 28 30

Completion time (Cj) 8 18 34 62 92

Therefore, the job sequence that minimises the mean flow time is 2-3-5-4-1.

Computation of minimum flow time (F min)

The flow time is the amount of time the job ‘f’ spends in the system. It is a measure which indicates thewaiting of jobs in the system. It is the difference between the completion time (Cj) and ready time (Rj)for job j.

Fj = Cj - Rj

Since the ready time (Rj) = 0 for all j, the mean flow time ( jF )is equal to Cj for all j. Mean flow time

( F ) = n

jj 1

1 1 1F [8 18 34 62 92] [214] 42.8

n 5 5=

= + + + + = =∑ hours

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(b) The weights are given as follows:

Job (j) 1 2 3 4 5Processing time (tj) hrs 30 8 10 28 16Weight (Wj) 1 2 1 2 3

Job (j) 1 2 3 4 5Processing time (tj hrs) 30 8 10 28 16Weight (Wj) 1 2 1 2 3tj/Wj 30 4 10 14 5.31

Job (j) 2 5 3 4 1


Fj = (Cj – Rj) 8 24 34 62 92

Wj 2 3 1 2 1

Fj x Wj 16 72 34 124 92

The weighted processing time = j

J

Processin g time(t )

Weight(W )

The weighted processing time is represented as

Thus, arranging the jobs in the increasing order of tj/Wj (weighted shortest processing time WSPT).

We have the optimal sequence that minimises the weighted mean flow time is 2-5-3-4 -1

flow time( wF ) :

wF =

∑

∑

n

j jj=1

n

jj=1

w F

w

The weighted mean flow time is computed as follows for optimal sequence.

Weighted mean flow time ( wF ) is computed as

wF = (16 72 34 124 92)

(2 3 1 2 1)+ + + +

+ + + + = 37.55 hrs.

Problem: 5The processing times and due dates of jobs for a single machine scheduling is given.

Determine the sequence which will minimise the maximum lateness and also determine the maximumlateness with respect to the optimal sequence.

Job (j) 1 2 3 4 5 6 7

Processing time (tj) 10 8 8 7 12 15 18

Due date (dj) 15 10 12 11 18 25 30


149

Solution:

Due date (dj) is the time at which job ‘j’ is to be completed and Lateness (Lj) is the amount of time bywhich the completion time of the job j differs from the due date (Lj = Cj - dj). Lateness is a measurewhich gives set of due dates of times. Lateness can be positive or negative. Positive lateness indicatesthe completion of job after its due date. It is therefore often desirable to optimize positive lateness.

Arranging the jobs as per Earliest Due Date (EDD), i.e., increasing order of their due dates. The sequenceis 2-4-3-1-5-6-7. This sequence gives the minimum value of maximum lateness (Lmax)

Computation of Lmax

Job (j) 1 2 3 4 5


Due date (in days) (dj) 16 20 25 15 40

From the above table, the maximum Lj is 48. This is the (optimised) value of Lmax.

Problem: 6

The processing times for five jobs and their due dates are given for a single machine scheduling below.

Job (j) 2 4 3 1 5 6 7as per EDD sequence

Processing time (tj) 8 7 8 10 12 15 18

Completion time (Cj) 8 15 23 33 45 60 78

Due date (dj) 10 11 12 15 18 25 30

Lateness (Lj) (-2) 4 11 18 27 35 48

(a) Determine the sequence

(b) Total completion time

(c) Average completion time

(d) Average number of jobs in the system and average job lateness using the following priority sequencingrules

(i) Shortest Processing Time (SPT)

(ii) Earliest Due Date (EDD)

(iii) Longest Processing Time (LPT)

(e) Compare the above characteristics for the three sequencing rules.

Solution:

(i) Shortest Processing Time (SPT) sequence.

As per this rule, the job with the shortest processing time is scheduled first and immediately followedby next lowest processing time and so on.

150


Job Sequence Processing time Flow time Due Date Job lateness(j) (tj) days (Fj) days (dj) days (Days)

3 5 5 25 0

5 6 11 40 0

2 7 18 20 0

1 9 27 16 11

4 11 38 15 23

Total 38 99 34

The various characteristics are:

Total completion/ time (flow time)= 38 days

Average completion time =Total flow time

No. of jobs = 995

= 19.8 days

Average number of jobs in the system = Total flow time 99

2.61 jobsTotal process time(completion) 38

= =

Average job lateness = Total Job lateness 34

No.of jobs 5= = 6.8 days

(ii) Earliest Due Date(EDD) rule

As per this rule priority is given to the job with earliest due date. Arranging the jobs as per EDDsequence gives the sequence as 4-1-2-3-5.

Characteristics:

• Total completion time = 38 days

• Average completion time = Flow timeNo. of jobs =

1285

= 25.6 days

• Average number of jobs in the system = Flow time

Completion time = 12838

= 3.37 jobs

• Average job lateness = 18

3.6days5

=

Job sequence Processing time Flow time Due Date Job lateness(j) (Cj) (Fj) (Dj) (Days)

4 11 11 15 0

1 9 20 16 42 7 27 20 73 5 32 25 75 6 38 40 0

Total 38 128 18


151

(iii) Longest Processing Time (LPT) Rule

The job sequence and computations as per this priority sequencing rule is given as follows:

Job Sequence Processing time Flow time Due Date Job lateness

(j) (Cj) (Fj) (Dj) (Days)

4 11 11 15 0

1 9 20 16 4

2 7 27 20 7

5 6 33 40 0

3 5 38 25 13

Total 38 129 24

Characteristics:

• Total completion time = 38 days

• Average completion time = Flow timeNo. of jobs =

1295

= 25.8days

• Average number of jobs in the system = Flow timeNo. of jobs =

12938

= 3.39 jobs

• Average job lateness = 245

= 4.8 days

(b) Comparison of Priority rules

Problem: 7

The following jobs are waiting to be processed in a turning shop today (July, 23). The estimates of the timeneeded to complete the jobs are as follows:

Jobs (j) Due date Processing time (tj) in days

1. July, 31 9

2. August, 2 6

3. August, 16 24

4. July, 29 5

5. August, 30 30

Sequence the jobs based on the minimum critical ratio.

Priority Total completion Avg. completion Avg. No. of jobs Avg. Jobrule time (days) time (days) in the system latenessSPT 38 19.8 2.61 6.8

EDD 38 25.6 3.37 3.6

LPT 38 25.8 3.39 4.8

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Solution:

The critical ratio is computed as

Critical Ratio (CR) = Time neededfor due dateof the job Time remaining

Time neededtocomplete the job Work remaining=

Critical Ratio (CR) =

As per the critical ratio rule, a job with the minimum critical ratio is given the first preference, i.e., thelower is the critical ratio, higher is its priority.

The denominator of CR, i.e., the time needed to complete the job includes the processing time remainingplus the transfer times plus the estimated waiting times remaining for the job to go through before itis completely processed.

The table below gives the calculations of time remaining and time needed and the critical ratio.

Jobs Due date Processing time Time needed to Critical ratio(j) (tj) in days complete the job in CR = Tr/Tn

days

1. July, 31 8 9 8/9 = 0.89

2. August, 2 10 6 10/6 = 1.167

3. August, 16 24 24 24/24 = 1.00

4. July, 29 6 5 6/5 = 1.20

5. August, 30 38 30 38/30 = 1.27

Note: Critical Ratio (CR) of less than one means that the job is already late.

The CR value of one indicates that the job is on schedule and greater than one indicates that the jobhas some slack available to it. From the table, job 1 has the lowest critical ratio and has to be processedfirst and job 2 has the highest critical ratio and it is scheduled last. The sequence is:

1 3 4 5 2

Problem: 8

A company has 8 large machines which receive preventive maintenance. The maintenance team is dividedinto 2 crews A and B. Crew A takes the machine power and replaces parts as per given maintenanceschedule. The second crew resets the machine and puts back into operation.

At all times the no passing rule is considered to be in effect. The servicing time for each machine is givenbelow.

Machine a b c d e f g h

Crew A 5 4 22 16 15 11 9 4

Crew B 6 10 12 8 20 7 2 21


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Determine the optimal sequence of scheduling the factory maintenance crews to minimise their idletimes and represent it on the Gantt chart.

Solution:

Step I : The minimum processing time on crew A is kept first in sequence while machine with processingtimes minimum on crew B is kept last in sequence.

Optimal sequence:

b h a e c d f g

Step II: Calculation of total elapsed time

WeekProduct 1 2 3 4

A 2000 4000 1000 2000B 3000 1000 4000 3000

Optimal Crew A Crew B Idle time on

Sequence S F S F Crew B

b 0 4 4 4 4

h 4 8 14 35 0

a 8 13 35 41 0

e 13 28 41 61 0

c 28 50 61 73 0

d 50 66 73 81 0

f 66 77 81 88 0

g 77 86 88 90 0

Crew B b h a e d f g

idle time

10 20 30 40 50 60 70 80 90 100Time in hrs

Total elapsed time is 90 hrs.

Gantt Chart

Problem: 9

The following is a tentative master schedule for 4 weeks at a small company

Crew a b h a e c d f g

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The labour in the key work centers for the company’s two major products is as follows

Product

Dept. A B

4 0.21 0.07

11 0.06 0.1

14 0.11 0.08

Determine the load on department 4 over the next 4 weeks.

Solution:

Hours required in each week:

Week 1: For product A = 0.21(2000) = 420 For product B = 0.07(3000) = 210 Total load for the week = 630 Week 2: For product A = 0.21(4000) = 840 For product B = 0.07(1000) = 70 Total load for the week = 910 Week 3: For product A = 0.21 (1000) = 210 For product B = 0.07(4000) = 280 Total load for the week = 490 Week 4: For product A = 0.21(2000) = 420 For product B = 0.07(3000) = 210 Total load for the week 630

Note that this load profile is not very uniform, even though 5000 units of products are produced eachweek.

Line Balancing

Assembly line balancing is associated with a product layout in which products are processed as they passthrough a line of work centres. An assembly line can be considered as a “production sequence” whereparts are assembled together to form an end product. The operations are carried out at different workstationssituated along the line.

The Problem of Line Balancing:

It arises due to the following factors

1. The finished product is the result of many sequential operations.

2. There is a difference in production capacities of different machines Line balancing is the apportionmentof sequential work activities into workstations in order to gain a high utilisation of labour andequipment so as to minimise the idle time. For example, the production capacities of two machines Aand B are as under for a particular job: A 50 pieces/hour; B 25 pieces/hour.


155

Now, if only one machine of each is provided, then machine B will produce 25 units/hour where asthe machine A can produce 50 units. But because of the sequence, only 25 units are produced perhour, i.e., machine A will work only 50 per cent of its capacity and the remaining 30 minutes in onehour, it is idle. This idle time can be minimised by introducing one more machine of kind B in theproduction line.

Steps in Solving Line Balancing Problems

1. Define task.

2. Identify precedence requirements.

3. Calculate minimum number of workstations required to produce desired output.

4. Apply heuristics to assign task to each station.

5. Evaluate effectiveness and efficiency.

6. Seek further improvement.

Three important parameters in line balancing

Total station time

1. Line efficiency (LE) = ××

Total station tim100

Cycle time no of workstations

2. Balance delay (BD) =Total idle time for all workstation

100Total available working time on all station

×

BD = (1-LE)

3. Smoothness Index (SI) = =∑

k

i l

(Max. station time - station times of station i)2

SI = 0, means a perfect balance

K = total number of workstations < total number of elements

Also, CT > maximum time of any work element n

Terms used in the context of Line Balancing:

1. Workstation: A work station is a location on assembly line where given amount of work is performed.

2. Cycle time: It is the amount of time for which a unit that is assembled is available to any operator onthe line or it is the time the product spends at each work station.

Cycle Time (CT) = Available time period AT

Output unitsrequired/period Output=

3. Task: The smallest grouping of work that can be assigned to a workstation.

4. Predecessor task: A task that must be performed before performing another (successor) task.

5. Task time: Standard time to perform element task.

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6. Station time (sk): Total standard work content of specific workstation.

7. Balance Delay (BD): Percentage of total idle time on the line to total time spent by the product frombeginning to end of line.

B.D =

n

k l

n.CT sk 100

n CT=

− ×

×

∑

B.D= Balance delay

n = number of work stations, CT = Cycle time

sk = Station time

Sequencing Model

This model is used for sequencing the jobs on the machines so as to minimize the total processing time. Wehave different sequencing models. They are:

(i) Sequencing ‘n’ jobs on 2 machines,

(ii) Sequencing ‘n’ jobs on 3 machines,

(iii) Sequencing ‘n’ jobs on ‘m’ machines,

(iv) Sequencing 2 jobs on ‘w’ machines,

(v) Sequencing of ‘n’ jobs on two machines.

The order in which jobs pass through machines or work-stations is a sequencing problem and analysis ofthis problem is sequence analysis. A production manager or a person who has been assigned the job ofsequencing the manufacturing of a product must know about the necessary operations and the possibleorder in which these can be performed so that die idle time can be minimised. The effectiveness of sequencingmay be measured by time and cost factors. Saving in time also affects the cost.

The main problem of sequencing is the assignment of various jobs to different machines. When there arevery few different types of jobs or machines, the problems is not much serious and it is solved informallyby sketching the flow mentally or on a time chart. Suppose, if these are two jobs and two machines (2x 2problem), there is no problem of assignment of jobs to machines. There are two possible sequence in thiscase —job 1 first and job 2 second or vice versa.

It becomes more tedious as the number of jobs and machines increases. We shall discuss or analyse thefollowing two cases:

(i) n jobs are to be processed on two machines A and B in the order AB.

(ii) n jobs are to be processed on three machines A, B and C in the order ABC.

(1) Assignment of jobs to two machines: Assuming that there are a number of jobs to be processedon two machines and the processing time for each job on each machine is known and that eachjob is first processed on Machine A and then on Machine B, the sequence of different jobs onmachine A and B may be assigned as follows:


157

(i) First, we should find out the least processing time. If it is on Machine A, it should be ranked first onMachine A. If it is on Machine B, it should be placed at the last in the sequence.

(ii) The next step would be to delete the job selected in step (i) above either to Machine A or B from the listof jobs to be assigned. This process should be repeated to the remaining jobs. This process will becontinued until a complete sequence of all the jobs is obtained. If two jobs have the same time on thesame machine (if it is smaller than the time on the other machine), the order of assignment for thesetwo jobs is arbitrary.

This technique applies to single unit or single type batch jobs where jobs have no priority for completion.In this technique it is also presumed that there is sufficient in-process storage space and the cost of in-process inventor is the same or varies very insignificantly for all units. These assumptions are valid onlyfor short processes. For extended process, oilier criterion such as closer inventor, cost control or expeditingpriorities etc. are considered other than minimising elapsed time. Other complicating variables may bevariable transportation time between machines, improving the defective work, breakdowns of machineand variable time caused by operator proficiencies or working conditions.

(2) Assignment of jobs to three machines: Assuming that there are a number of jobs to be processed onthree machines A, B and C in the order of ABC, i.e., first it is processed on machine A, then onMachine B and then on machine C. The problem can be solved if either of the two conditions issatisfied.

(i) The least processing time of any job on Machine A is greater than or equal to the maximumprocessing time of any job on Machine B.

(ii) The smallest processing time for any job on Machine C is either equal to or greater than theprocessing time for any job on Machine B.

If any of the above two conditions is satisfied, the job processing times are converted into the followingways:

a. We shall add job processing times (as given) of different jobs on; Machine A and B; and

b. Again job processing time for various jobs (given) on Machine B and C shall be added.

After this process, the optional sequence can be obtained in the same way as it is assigned on two machines.

2.9 Economic Batch Production

Production managers often have to decide what quantity of output must be produced in a batch (known aslot size or batch size). The products are manufactured in lot sizes against the anticipated demand for theproducts. Often the quantity produced may exceed the quantity which can be sold. (i.e., production ratesexceed demand rates). The optimum lot size which is known as economic lot size or economic batch quantityor economic manufacturing quantity is that quantity of output produced in one batch, which is mosteconomical to produce, i.e., which results in lowest average cost of production.

Determination of Economic Lot Size for Manufacturing:

The factors to be considered in arriving at the economic lot size are:

(i) Usage rate: The rate of production of parts should match with the rate of usage of these parts in theassembly line.

158


(ii) Manufacturing cost: Higher the lot size, lower will be the cost per unit produced because of distributionof set up costs for setting up production or machines and preparing paper work (production orders).But the carrying cost (handling and storing costs) will increase with increase in lot size.

(iii) Cost of deterioration and obsolescence: Higher the lot size, higher will be the possibility of loss dueto deterioration (items deteriorating after shelf life) or obsolescence (due to change in technology orchange in product design).

Before deciding on production using economic lot sizes, the availability of production capacity to producethe product in economic lot size must be verified. The economic lot size balances the two opposing costsrelated to batch size i.e., setup cost for production and the inventory carrying costs resulting from inventoryof products produced when production rate exceeds usage rate or when the items produced are notimmediately consumed in the next stage of production. The set up cost per unit decreases with increase inlot size whereas the inventory carrying cost increases with increase in lot size. Diagram below illustratesthe concept of economic batch quantity or economic lot size.

If S is the set up cost per set up, ‘C’ is the production cost per unit produced and I is the inventory carryingchanges (%) and A is the annual demand for the item in units, then,

Economic Batch Quantity (EBQ)

or Economic Lot Size (ELS)

or Economic Manufacturing Quantity (EMQ)

⎛ ⎞ ⎛ ⎞× ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= =

⎛ ⎞ ⎛ ⎞×⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Annualdemand Set upCost2

(in units) per set up2ACProduction Cost Inventory conveyingCI

per unit ch arg es (percentage)


159

Economic Run Length: When a firm is producing an item and keeping it in inventory for later use, insteadof buying it, the formula used to calculate economic order quantity (EOQ) can be used to calculate theeconomic production quantity referred to as Economic Run Length (ERL).

If ‘p’ is the production rate and ‘d’ is the demand rate (or consumption rate), A is the annual demand forthe item in units, I is the inventory carrying charges (percentage), C is the production cost per unit, then,

Economic Run Length (ERL) =

2AS

dCI 1 –

p⎛ ⎞⎜ ⎟⎝ ⎠

=

⎛ ⎞ ⎛ ⎞× ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞× − ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Annualdemand Set upCost2

(in units) per set upProduction cos t Inventory conveying DemandRate

per unit ch arg es (percentage) ProductionRate


Problem: 1

C Ltd. produces a product which has a monthly demand of 4,000 units. The product requires a componentX which is purchased at Rs. 20. For every finished product, one unit of component is required. The orderingcost is Rs. 120 per order and the holding cost is 10% p.a.

You are required to calculate:

(i) Economic order quantity.

(ii) If the minimum lot size to be supplied is 4,000 units, what is the extra cost the company has toincur?

(iii) What is the minimum carrying cost, the company has to incur.

Solution:

(i) Economic Order Quantity:

Annual consumption of component X (C0):

4,000 units per month for 12 months = 48,000 units;

Ordering cost per order (O) = Rs. 120 and Carrying cost (i.e., holding cost) of one unit of componentfor one year (Cc) = 10% of Rs. 20 = Rs. 2

EOQ = 2 48,000 units Rs.1202

2, 400 unitsRs.2

× ×= =O

O

C OC

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(ii) Extra cost to be incurred by the company:

Cost when order size is 4,000 units ⎛ ⎞⎜ ⎟⎝ ⎠

48,000i.e., or 12 orders p.a.

4,000

Rs.

Ordering cost : 12 order p.a. at Rs. 120 per order 1,440

Carrying cost of average inventory :

4,000 units

Rs.22

× 4,000

Total annual cost 5,440

cost when order size is 2,400 units (i.e, 48,0002, 400 or 20 orders p.a.)

Rs.

Ordering costs : 20 orders p.a. at Rs.. 120 per order 2,400

Carrying cost of average inventory: 2,400

Total annual cost 4,800

Extra cost to be incurred : Rs. (5,440 - 4,800) = Rs. 640.

(iii) Minimum carrying cost to be incurred by the company:

Since carrying cost depends upon the size of the order, carrying cost will be the minimum at theeconomic order quantity. Hence the minimum carrying cost is Rs. 2,400 as calculated in (ii) above.

Problem: 2

M/s Kobo Bearings Ltd., is committed to supply 24,000 bearings per annum to M/s Deluxe Fans on asteady daily basis. It is estimated that it costs 10 paisa as inventory holding cost per bearing per month andthat the setup cost per run of bearing manufacture is Rs. 324.

(a) What is the optimum run size for bearing manufacture? (b) What should be the interval between theconsecutive optimum runs? (c) Find out the minimum inventory holding cost.

Solution:

(a) Optimum run size or Economic Batch Quantity (EBQ)

=× × × ×= =

×2 2 24000 324

3600units0.10 12

AnnualOutput SetupcostAnnualCost of Carrying one unit

(b) Interval between two consecutive optimum runs = 30×EBQMonthly Output


161

= 3600

30 5424000 12

× =÷ Calendar days

(c) Minimum inventory holding cost = Average inventory × Annual carry-ing cost of one unit ofinventory

= (3600÷2) ×0.10×12 = Rs. 2,160.

Problem :3

The demand of a certain item is random. It has been estimated that the monthly demand of the item has anormal distribution with a mean of 680 and a standard deviation of 130 units. The unit price of the item isRs. 10 per unit; the ordering cost is Rs. 20. The inventory carrying cost is estimated to be 25 per cent peryear respectively. The procurement lead time is constant and is one week. Find the most economic orderingpolicy and the expected total cost of controlling inventory, given that the service level is 97.5%.

Solution:

Under the given circumstances, Economic Order Policy involves deter-mination of EOQ, ROL andSafety Stock for a year.

EOQ = 2 680 12 2

0.25 10× × ×

× = 361 (appx.) units; ROL = Consumption during lead time + Safety stock

= 680 ÷ 4 + ZLT. Z for 97.5% service level may be taken as 2 (1.96 from Table)

LT = S.D. during lead time = Varience During Lead time = 16900

4 = 130/2 = 65 units

Safety stock = 2 × 65 = 130 units; ROL = 680/4 + 130 = 300 units

Expected total cost of inventory per annum: Procurement cost = Rs. 10 × 680 × 12 = Rs. 81,600

Ordering cost = (Annual Demand / EOQ) × (Ordering cost / Order) = (680 × 12 × 20)/361.

= Rs. 452 (apx.)

Inventory carrying cost = Average inventory × inventory carrying cost = (EOQ/2 + Safety stock) ×0.25 × Rs. 10 = (361/2 + 130) × 0.25 × Rs. 10 = Rs. 776 (apx.) Total variable cost = Rs. 452 + Rs. 776 = Rs.1,228 .

Problem: 4

A manufacturer requires 10,00,000 components for use during the next year which is assumed to consist of250 working days. The cost of storing one component for one year is Rs. 4 and the cost of placing order isRs. 32.

There must always be a safety stock equal to two working days usage and the lead time from the supplier,which has been guaranteed, will be five working days throughout the year. Assuming that usage takesplace steadily throughout the working days, delivery takes place at the end of the day and orders areplaced at the end of working day, you are required to(a) Calculate the EOQ (b) Calculate the Re-orderpoint.

Solution:

We are given that :- D = 10,00,000 ; S = Rs. 32; C = Rs.4

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EOQ = 2DS 2(10,00,000)32

C 4= = 4000 units.

(b) Re-order point is to be calculated by the following formula;

Re-order point = Safety stock + (Average usage × Lead time)

Safety stock = (10,00,000 / 250days) = 4,000 usage per day × 2 = 8,000.

Reordering point = 8000 + (4000 units per day × Number of days)

= 8000 + (4000 ×5) = 28,000 units.

Problem: 5

The monthly requirement of raw material for a company is 3000 units. The carrying cost is estimated to be20% of the purchase price per unit, in addition to Rs. 2 per unit. The purchase price of raw material is Rs. 20per unit. The ordering cost is Rs. 25 per order. (i) You are required to find EOQ.(ii) What is the total costwhen the company gets a concession of 5% on the purchase price if it orders 3000 units or more but lessthan 6000 units per month. (iii) What happens when the company gets a concession of 10% on the purchaseprice when it orders 6,000 units or more? (iv) Which of the above three ways of orders the company shouldadopt?

Solution:

We are given that,

D = 3,000 × 12 = 36,000 units per annum ; S = Rs. 25;

C = 2 + 20% of Rs. 20 = 2 + 4 = Rs. 6

(i) EOQ = 2DS 2 36000 25

300,000C 6

× ×= = = 548 units (approx.)

Total cost= Ordering Cost + Cost of raw material + Storage cost

= (36,000 / 548) × 25 + (36,000 × 20)+(548/2) ×6

= Rs. 1642.33 + 7,20,000+ 1,644 = Rs. 7,23,286.

(ii) When the company has an option to order between 3000 and 6000 units, the EOQ should becalculated with a reduction in price by 5% (due to concession); The purchase price = 95% of Rs.20 = Rs. 19.

D = 36,000 units per annum; S = Rs. 25; C = 2 + 20% of 19 = 2 + 3.80 = Rs 5.80

EOQ = 2 36000 25 18,00,000

5.80 5.80× × = = 557 units app.

Total cost = (36,000/557)× 25+(36,000 × 19)+ (557/2) ×5.80

= Rs. (1,615.79 + 6,84,000+1,615.30) = Rs. 6,87,231.09

(iii) When the company orders more than 6,000 units purchase price = 90% of Rs. 20 (because 10%concession)


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= Rs. 18; D = 36,000 units per annum; S = Rs. 25; C = 2 + 20% of Rs. 18

= 2 + 3.60 = 5.60

EOQ = 2DS 2 36000 25

C 5.60× ×= = 567 units app. Total cost

= (36,000/567) ×25+ (36,000×18)+ (567/2) ×5.60

= Rs. 1,587.30 + 6,48,000 + 1,587.60 = Rs. 6,51,174.90

(iv) Comparing these costs, we notice that the cost is minimum for (iii) order. Therefore the companyshould adopt a policy of ordering 567 units per order.

Problem: 6

M/s. Tubes Ltd. are the manufacturers of picture tubes of T.V. The following are the details of their operationduring 2001:

Average monthly market demand 2,000 tubes

Ordering cost Rs. 100 per order

Inventory carrying cost 20% per annum

Cost of tubes Rs. 500 per tube

Normal usage 100 tubes per week

Minimum usage 50 tubes per week

Maximum usage 200 tubes per week

Lead time to supply 6 – 8 weeks

Compute from the above:

(1) Economic order quantity. If the supplier is willing to supply quarterly 1,500 units at a discountof 5%, is it worth accepting?

(2) Maximum level of stock.

(3) Minimum level of stock.

(4) Re-order level of stock.

Solution:

(1) Economic Order Quantity:

Annual usage of tubes (Co) = Normal usage per week × 52 weeks

= 100 tubes × 52 weeks

= 5,200 tubes.

Ordering cost per order (O) = Rs. 100.

Inventory carrying cost per unit per annum = 20% of Rs. 500 = Rs. 100.

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EOQ = 2 5, 200 units Rs.100

Rs.100× ×

=O

O

C OC = 102 units (approx.).

Evaluation of order size of 1,500 units at 5% discount

No. of orders = 5, 200 units1, 500 units = 3.46 units or 4 (in case of a fraction, the next whole number is considered).

Rs.

Ordering cost order per year at Rs. 100 per order 400

Carrying cost of average inventory:

1, 500 units 20

Rs.(500 less5%)2 100

× × 71,250

Total annual cost (excluding item cost) 71,650

Annual cost if EO Q (102 units) is adopted : Rs.

Ordering cost: 5,200 ÷ 102 or 51 orders per year of Rs. .100 per order 5,100

Carrying cost of average inventory 102 units 20

Rs.5002 100

× × 5,100

Total annual cost (excluding item cost) 10,200

Increase in annual cost: Rs. (71,650 – 10,200) = Rs. 61,450.

Amount of quantity discount: 5% × Rs. 500 × 5,200, units = Rs. 1,30,000.

Since the amount of quantity discount (Rs. 1,30,000) is more than the increase in total annual cost(Rs. 61,450), it is advisable to accept the offer. This will result in a saving of Rs. (1,30,000 - 61,450) orRs. 68,550 p.a. in inventory cost.

(2) Maximum Level of Stock:

= Re-order level + Re-order quantity – (Minimum usage × Minimum delivery period)

= 1,600 units + 102 units – (50 units × 6 weeks) = 1,402 units.

(3) Minimum Level of Stock:

= Re-order level – (Normal usage × Normal delivery period) [see Note]

= 1,600 units – (100 units × 7 weeks) = 900 units.

Note: Normal delivery period is taken to be the average delivery period.

(4) Re-order Level of Stock:

= Maximum usage × Maximum delivery period = 200 units × 8 weeks = 1,600 units.


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2.10 Human Resource Planning

HRP is a process of stricking balance between human resources required and acquired in an organisation.In other words, HRP is a process by which an organisation determines how it should acquire its desiredmanpower to achieve the organisational goals. Thus, HRP helps an organisation have the right numberand kind of people at the right places and right times to successfully achieve its overall objectives. Humanresource planning is a process of determining and assuming that the organisation will have an adequatenumber of qualified persons, available at the proper times, performing jobs which meet the needs ofenterprise and which provide satisfaction for the individuals involved.

HRP is the process- including forecasting, developing and controlling-by which a firm ensures that it hasthe right number of people and the right kind of people at the right places at the right time doing work forwhich they are economically most useful.

HRP can be defined as the comparisons of an organisation’s existing labour resources with forecast labourdemand, and hence the scheduling of activities for acquiring, training, redeploying and possibly discardinglabour. It seeks to ensure that an adequate supply of labour is available precisely when required.

HRP could be seen as a process, consisting of the following series of activities:

1. Forecasting future personnel requirements, either in terms of mathematical projections of trends inthe economy and developments in the industry, or of judgements estimates based upon specific futureplans of the company.

2. Inventing present manpower resources and analysing the degree to which these re-sources areemployed optimally.

3. Anticipating Manpower Problems by projecting present resources into the future and comparingthem with the forecast of requirements, to determine their adequacy, both quantitatively andqualitatively.

4. Planning the necessary programmes of recruitment, selection, training, employment, Utilisation,transfer, promotion, development, motivation and compensation so that future man powerrequirements will be duly met.

Objectives of HRP :

The main objective of having human resource planning is to have an accurate number of employees required,with matching skill requirements to accomplish organisational goals. In other words, the objectives ofhuman resource planning are to:

• Anticipate the impact of technology on jobs and requirements for human resources.

• Assess surplus or shortage, if any, of human resources available over a specified period of time.

• Control the human resources already deployed in the organisation.

• Ensure adequate supply of manpower as and when required.

• Ensure proper use of existing human resources in the organisation.

• Forecast future requirements of human resources with different levels of skills.

• Provide lead time available to select and train the required additional human resource over a specifiedtime period.

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Importance of HRP:

1. Despite growing unemployment, there has been shortage of human resources with required skills,qualification and capabilities to carry on works. Hence the need for human resource planning.

2. Human resource planning is also essential in the face of marked rise in workforce turnover which isunavoidable and even beneficial. Voluntary quits, discharges, marriages, promotions and seasonalfluctuations in business are the examples of factors leading to workforce turnover in organisations.These cause a constant change and flow in the work force in many organisations.

3. Human resource planning is also needed in order to meet the needs of expansion and diversificationprogrammes of an organisation.

4. Large numbers of employees, who retire, die, leave organisations, or become incapacitated because ofphysical or mental ailments, need to be replaced by the new employees. Human resource planningensures smooth supply of workers without interruption.

5. Technological changes and globalisation usher in change in the method of products and distributionof production and services and in management techniques. These changes may also require a changein the skills of employees, as well as change in the number of employees required. It is human resourceplanning that enables organisations to cope with such changes.

6. The need for human resource planning is also felt in order to identify areas of surplus personnel orareas in which there is shortage of personnel. Then, in case of surplus personnel, it can be redeployedin other areas of organisation. Conversely, in case of shortage of personnel, it can be made good bydownsizing the work force.

Human resource planning is important to organisation because it benefits the organisation in several ways.The important ones are mentioned below:

1. By maintaining a balance between demand for and supply of human resources, human resourceplanning makes optimum use of human resources, on the one hand, and reduces labour costsubstantially, on the other.

2. Careful consideration of likely future events, through human resource planning might lead to thediscovery of better means for managing human resources. Thus, foreseeable pitfalls might be avoided.

3. Human resource planning compels management to assess critically the strength and weaknesses ofits employees and personnel policies on continuous basis and, in turn, take corrective measures toimprove the situation.

4. Human resource planning helps the organisation create and develop training and succes-sion planningfor employees and managers. Thus, it provides enough lead time for internal succession of employeesto higher positions through promotions.

5. Human resource planning meets the organisation need for right type of people in right number atright times.

6. It also provides multiple gains to the employees by way of promotions, increase in emolu-ments andother perquisites and fringe benefits.

7. Last but no means the least, with increase in skill, knowledge, potentialities, productivity and job satisfaction,organisation becomes the main beneficiary. Organisation is benefited in terms of increase in prosperity /production, growth, development, profit and, thus, an edge over its competitors in the market.


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8. Manpower shortfalls and surpluses may be avoided, to a large extent.

9. Some of the problems of managing change may be foreseen and their consequences mitigated.Consultations with affected groups and individuals can take place at an early stage in the changeprocess. This may avoid resistance for change.

10. Through human resource planning, duplication of efforts and conflict among efforts can be avoided,on the one hand, and coordination of worker’s efforts can be improved, on the other.

Forecast requirement for human resources in the future:

There are various techniques varying from simple to sophisticated ones employed in human resourceforecasting. These include:

1. Management Judgement,

2. Work-Study Method,

3. Ratio-Trend Analysis,

4. Delphi Technique,

5. Flow Models,

6. Mathematical Models.

2.11 Material Requirement Planning

Material requirement planning (MRP) refers to the basic calculations used to determine componentrequirements from end item requirements. It also refers to a broader information system that uses thedependence relationship to plan and control manufacturing operations.

MRP is a technique of working backward from the scheduled quantities and needs dates for end itemsspecified in a master production schedule to determine the requirements for components needed to meetthe master production schedule. The technique determines what components are needed, how many areneeded, when they are needed and when they should be ordered so that they are likely to be available asneeded. The MRP logic serves as the key component in an information system for planning and controllingproduction operations and purchasing. The information provided by MRP is highly useful in schedulingbecause it indicates the relative priorities of shop orders and purchase orders.

“Materials Requirement Planning (MRP) is a technique for determining the quantity and timing for theacquisition of dependent demand items needed to satisfy master production schedule requirements.”

MRP is one of the powerful tools that, when applied properly, helps the managers in achieving effectivemanufacturing control.

MRP Objectives:

1. Inventory reduction: MRP determines how many components are required, when they are requiredin order to meet the master schedule. It helps to procure the materials/components as and whenneeded and thus avoid excessive build up of inventory.

2. Reduction in the manufacturing and delivery lead times: MRP identifies materials and componentquantities, timings when they are needed, availabilities and procurements and actions required to

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meet delivery deadlines. MRP helps to avoid delays in production and priorities production activitiesby putting due dates on customer job orders.

3. Realistic delivery commitments: By using MRP, production can give marketing timely informationabout likely delivery times to prospective customers.

4. Increased efficiency: MRP provides a close coordination among various work centres and hencehelps to achieve uninterrupted flow of materials through the production line. This increases theefficiency of production system.

Functions served by MRP

1. Order planning and control: When to release orders and for what quantities of materials.

2. Priority planning and control: How the expected date of availability is compared to the need date foreach component.

3. Provision of a basis for planning capacity requirements and developing a broad business plans.

2.12 Productivity Measurement Techniques of Factors of Production

Productivity implies development of an attitude of mind and a constant urge to find better, cheaper, quicker,easier and safe ways of doing a job manufacturing an article and providing a service. Since the beginningof the industrial era, the manufacturers or producers have been facing the problem of how to use theavailable resources and factors of production to the best of their ability and capacity so as to get the maximumoutput with the minimum cost of production. Industrial revolution, social, technological and scientificdevelopments, changes in economic systems is the various efforts made in this direction and the process ofdevelopment and changes is still on. New and new machines, methods and technology are being inventedand used in the industrial field to minimise the wastage of men, materials and machines. It is all to increasethe productivity.

Productivity is the quality or state of being productive. It is some relationship of outputs to inputs. It is aconcept that guides the management of a production system, and measures its success. It is the quality thatindicates how well labour, capital, materials and energy are utilised. Productivity improvement is soughteverywhere because it supports a higher standard of living, helps control inflation, and contributes towardsa stronger national economy.

Productivity is an indicator reflecting the changes in the performance of the enterprise and having somesort of input-output comparisons relating to various activities of an organisation. It also facilitates themanagement to control and plan its future operations of the enterprise.

Productivity is the talk of the day and it is generally regarded as efficiency in industrial production to bemeasured by some relationship between outputs and inputs. The increase in productivity is looked uponas the key to prosperity at all levels. In its modern sense, it refers to the relationship between the result andthe means employed or to be more specific between the product and the factors used for obtaining it. It isthe quantitative relationship between what we produce and (the resources which we use to obtain it. It canalso be termed as the ratio of what is produced to what is required to produce it. The higher is the ratio, thegreater is the productivity. Thus, it seeks to measure the economic soundness of the use of the means. Itmeans productivity can be considered higher if the same amount of production is obtained with lessermeans or it will be lower if the same quantity of produce is obtained with larger quantity of means. It is


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higher when there is maximum production with the least expense of resources.

A productivity index is a device of expressing the ratio between outputs and the inputs of the resourcesnumerically. These indices are prepared by comparing the volume of output of goods with the labouremployed on that job or the profits of the firm with the capital employed. If the comparison shows anupward trend in indices, it is a sign of improved or better productivity or vice-versa.

The productivity is a measure of how much input is required to achieve a given output.

Symbolically:

P = OI

where P = Productivity;

O = Output,

I = Input.

The output may be measured in terms of the units of goods produced or the value of goods and servicesproduced. The input, on the other hand, can be referred to as the combination of different factors, i.e., rawmaterials, machinery, worker’s time, power, efforts and imagination of entrepreneur and the managers. Aunit of input, therefore, can be expressed as one worker, or one hour of labour time or one tonne of rawmaterials, or one kw of electricity and so on. Thus, it is very clear from the above description that theproductivity can be calculated or measured for each one of the factors comprising the input or of all thefactors together. The productivity of labour, for example, can be found out by ascertaining the ratio betweenthe quantity of goods produced and the number of workers or man-hours employed on the production ofsuch output.

The importance of the concept of productivity can be viewed from the following points:

1. To beat the competition: It is an age of cut-throat competition. There may be other commoditieswhich can serve as the substitutes of the terms ‘product’ and can attract the consumers’ purchasingpower. The firm whose productivity is higher can only beat the competition and can exist in themarket for long.

2. Guide to Management: The productivity indices are very useful for the management and can be usedfor different purposes. These indices can serve as a valuable guide to the management for improvingthe performance of its enterprise. The productivity measures can be used for the following purposes:

(a) Strategic : With the help of productivity indices, the efficiency of different firms can be measured,analysed and compared. The necessary steps can be taken to improve the productiveness of thefirm taking in view the productiveness of the other competitive firms.

(b) Tactical: Different units or the sectors of the firm can also be compared as regards to theirproductivity and the productivity of the less productive units or sectors can be improved.

(c) Planning: A firm uses different inputs in producing the goods. A comparison of relative benefitsaccruing from the use of different inputs can be had and the most beneficial input can be used inproduction. It helps the management to plan for the future.

(d) Administration: Productivity indices indicate the progress of the firm over a period of years.The productivity of different inputs, including labour, can be measured individually. Theindividual productivity indices help the management in bargaining with the labour leaders,

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trade unions and the Government in case of labour disputes regarding welfare activities. Thusadministration can be improved with the help of productivity indices.

3. An Indicator of Progress: In economically backward countries, productivity movement is basic aspectof progress. It implies the development of an attitude of mind and a constant urge to better, cheaper,quicker and safer ways of doing a job, manufacturing a product and providing a service. In an urge toimprove the productivity, new inventions take place. This productivity is an aspect of basic progress.

4. Maximum utilisation of Scarce Resources: In order to provide the articles or commodities to theconsumers at the lowest possible cost, the productivity urges to utilise the available resources to themaximum to the satisfaction of customers. The productivity processes and techniques are designed tofacilitate more efficient work involving less fatigue to workers by improvements in the layout of theplant and work, better working environment and simplification of works.

5. Key to National Prosperity: The productivity, in fact, has become the synonymous to progress. Higherproductivity is an index of more production with the same inputs at lower cost. It enables industry tooffer goods to the general public at cheaper rates and results in expansion of markets. The workingconditions and wages of workers will improve and industrialists too will get larger profits. Thushigher productivity is the key to national prosperity. The secrets of Japan and Western countries’prosperity lie in increased productivity.

6. Prosperity to Labour: The higher productivity is a boon to labour also. It brings improved workingconditions, better wages and salaries to workers, better labour welfare activities to labourers. Thustheir standard of living is improved.

7. Other Uses:

(i) Higher productivity increases the profits and reserve funds of the industry that can be used forexpansion and modernisation.

(ii) It increases the goodwill of the firm due to cheaper goods to the public, well-off staff and moreprofits and better financial position.

(iii) It improves the competitive strength of the company in export markets through reduction in costof production and quality products.

In this way, productivity is the only way to make the overall progress of the country.

Measurement of Productivity: The productivity or the performance of various input and output factorscan be measured in many ways. These measures are mainly based on the following two criteria:

(i) Change in output per unit of input: indicates the change in the performance of corresponding inputduring the given period, e.g., change in output per worker or per man-hour will signify the change inperformance of labour.

(ii) Change m input per unit of output: during the given period signifies the change in the performanceof the corresponding input factor, e.g., change in man-hour or workers’ per unit of output will alsoindicate the change in the performance of the labour input.

Productivity measurement implies the use of standards set for each input factor in terms of output. Incircumstances where standards are not in use, productivity can be measured only when the output isconverted into ‘units or work’ which is defined as the amount of work that can be performed by one unit


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of input. Thus productivity can be measured by dividing the output by the performance of each inputfactor taken together.

Some of the well-known indices of productivity are given below:

(A) Man-hour output: The most widely used index of productivity is to work out the output per man-hour it can be put as –

Productivity = Units of output / Total man-hours

(B) Productivity Ratio : The rate of return on capital employed is a valuable and widely used guide tomany types of business decisions. This ratio of profit to capital employed is a valuable means ofmeasuring the performance of divisions, sections, plants, products and other components of a business,and can be calculated as—

Productivity = Net Profit / Capital employed

(C) Use of Financial Ratios: There are many situations when time standards cannot be set and therefore,it is very difficult in such cases to measure the productivity by a direct method. In these cases, financialratios can be used to measure the productivity by using its sales turn-over. But ‘added value’ is amore useful approach for measuring productivity. ‘Added value’ means output - inputs.

The most common financial ratio of productivity is—

Productivity = Added Value / Labour Costs

Productivity = Added Value / Conversion Costs

The first ratio gives the financial productivity of labour force and the second ratio gives the financialproductivity of all the resources of the company put together.

(D) Other Useful Measures: There are many other useful productivity ratios to measure the productivityof various input factors. These are:

(i) Manpower Productivity =Value of output of goods or servicesNo.of wor ker s or man hours used

(ii) Materials Productivity =Value of output of goodsor services

Units (or cos t)of materials used

(iii) Capital Productivity =Valueof outputofgoodsorservices

Capitalassetsemployed

(iv) Energy Productivity =Value of output of goodsor services

Unit(or cos t)of energy used

A combined measure of productivity can be taken as

Productivity = Valueof output of goodsor services

Valuesof (labour capital materials other in puts)+ + +

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There may be other input factors such as insurance, taxes, advertising etc. and their productivity can bemeasured likewise.

Each measure requires different kinds of data and only rarely such information is available for allcommodities in an industry on continuous basis.

Tools of productivity or how to increase productivity:

The productivity of an enterprise can be improved by improving the performance of various inputs andother factors affecting productivity. For this purpose, use of following tools can be recommended.

1. Human Aspects: Under this, cooperation of workers is sought in the following ways:

(i) More workers’ participation in management or in decision making through joint consultation.

(ii) Improving communication services.

(iii) Improving mutual trust and cooperation through improved job procedures, better training ofemployees, more workers incentives by implementing various incentive schemes, and labourwelfare programmes.

(iv) Better planning of work, more effective management, more democracy in administration,improved human relations and selection and training of personnel at various levels of managementare some human efforts from the side of management in order to improve the productivity.

2. Supply of Inputs:

(i) Improvement in the nature and quality of raw materials and their supplies to the work.

(ii) Proper provision of plant, equipment and their maintenance.

(iii) Introduction of more and more machines and equipment in place of physical work.

(iv) Fuller utilisation of manpower and efficiency or capacity of plant and equipment employed.

3. Technological Aspects:

Certain methodological and technological developments are also necessary to improve the productivity ofthe concern.

These are;

(i) Work, time and motion studies to determine better ways and means of doing a job.

(ii) Implementing various simplification, specialisation and standardisation programmes.

(iii) Applying control techniques comprising production and planning control, cost control and qualitycontrol techniques.

(iv) Improving layout of plants, shops and machine tools, and material handling and internaltransportation system.

(v) Improving inspection techniques so as to minimise the wastage and defective work.

Factors affecting industrial productivity:

Productivity is defined to be some ratio between output and input. Thus all factors which affect outputand inputs will also affect the measure of productivity.


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The following factors affect the productivity.

1. Technological Development: Technical factors including the degree of mechanisation, technicalknow-how, raw materials, layout and the methods and techniques of work determine the level oftechnological development in any industry. The principal factors in technological developmentaffecting productivity are:

(a) The Size of the Plant: The size of the plant and the capacity utilisation has direct bearing onproductivity. Production below or above the optimum level will be uneconomical and will tendtowards lower level of productivity.

(b) Research and Development: Investment in research and development may yield better methodof work and better design and quality of products.

(c) Plant and Job Layout: The arrangement of machines and positions in the plant and the set-up ofthe work-bench of an individual worker will determine, how economically and efficientlyproduction will be carried out.

(d) Machine and Equipment Design: Whether the design of machinery and equipment is modernand in keeping with the limitations and capacities of the workers will also determine theproduction efficiency and level of productivity.

(e) Production Processes: Advanced production processes involving the use of modern integratedand automatic machinery and semi-processed materials have been known to help in raisinglevels of productivity.

(f) Power, Raw Materials etc. Improved quality of raw materials and increased use of power havea favourable effect on productivity.

(g) Scientific Management Techniques: Scientific management techniques such as better planningof work, simplification of methods, time and motion study, emphasis for reduced wastage andspoilage have positive effects on productivity.

It will be realised that technological development requires a great amount of funds and generaleconomic and technical environment in the country. Thus capital plays an important role in increasingthe productivity through implementing technological development. It should also be recognised thatsuch developments influence the job performance of employees. With better machines, tools andprocesses, it should be considered that both ability and willingness to work should be increased.

2. Individual Factors: Individual factors such as knowledge, skill and attitude also affect the productivityof industry. Knowledge is acquired through training, education and interest on the part of learner.Skill is affected by aptitude (one’s capacity to learn a particular kind of work), personality (emotionalmaturity, balance of mind etc.) as also by education, experience, training etc. Increased knowledge,skill and aptitude certainly increased the productivity and a person deficient in these personal attributesis less productive than an average man.

The attitude (willingness of employee to work for organisation) of employees towards the work andthe organisation affect their productivity to a great extent. Knowledge and skill without willingnessare futile. The urge to work is a complex phenomenon governed by several factors such as formal andinformal organisation, leadership, need, satisfaction, influence of trade unions etc. These factorsmotivate the workers to work better and with enthusiasm.

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3. Organisation Factors: Organsiation factors include various steps taken by the organisation towardsmaintaining better industrial relations such as delegation and decentralisation of authority, participativemanagement (workers’ participation in management), organisational efficiency, proper, personnelpolicies relating to selection, placement, promotion, wage salary levels, incentives, merit rating, jobevaluation, training and provision for two-way communication, supervision, etc. These factors alsoinfluence motivation. Likewise the existence of groups with higher productivity as their goal is likelyto contribute to the organisational objectives. These facts were brought out by Hawthorne experimentsin U.S.A. A properly-motivated worker will certainly contribute to the industrial productivity.

4. Work Environment: The importance of proper work environment and physical conditions on the jobhas been emphasised by industrial psychologists and human engineers. Better work environmentensures the greatest ease at work through better ventilation and light arrangement, improved safetydevices, reduction in noise, introducing suitable rest-pause etc.

5. Other factors: There are several other factors that affect productivity. These are:

(a) Natural Factors: Physical, geographical and climatic conditions influence the productivity atlarge. Abundance of natural resources affects the productivity and similarly climate affects theefficiency of workers to a great extent.

(b) Managerial Factors: The industrial productivity is influenced very much through managerialability and leadership. The managerial ability of utilising the available resources to the maximum,organising capacity, foresightedness, decision-making ability and entrepreneurship are certainfactors that contribute to productivity.

(c) Government Policy: Government policies towards industry also contribute to industrialproductivity. Taxation policy, financial and administrative policy, tariff policy and protectionpolicy affect the productivity to a large extent.

Thus, the above factors are responsible for the increased productivity.

Production and Productivity:

Production and productivity are not synonymous. Production refers to the volume, value or quantity ofgoods and services produced during a given period by a worker, plant, firm or economy. It is the sum totalof results achieved by the various factors used together. Productivity, on the other hand, is not concernedwith the volume of production. It is the ratio of output and input factors of an enterprise. It shows theefficiency of production or the efficiency level of input factors. In other words, productivity is relative tothe resources used in turning out a certain amount of physical output, while production is used, more orless, in absolute sense. The distinction between (lie two terms becomes more clear when we find thatincrease in production does not necessarily mean the increase in productivity. If increase in production isattributed to the increase in the inputs of production in the same proportion, the production will haveincreased but productivity may have declined or may remain constant because the ratio of output andinputs has shown a decline or has not shown any improvement.


Problem: 1 In a particular plant there are 10 workers manufacturing a single product and the output permonth consisting of 25 days of that particular product is 200. How much is the monthly productivity?


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Solution:

Monthly productivity per worker = 20010

= 20 units

Problem: 2 There are two industries A and B manufacturing hose couplings. The standard time perpiece is 15 minutes. The output of two small scale industries is 30 and 20 respectively per shift of 8hours. Find the productivity of each per shift of 8 hours. What is the expected production of each perweek consisting of 6 days?

Solution:

Productivity = Actual production

S tan dard production

Standard production of hose complings per shift = 8 60

15×

= 32 pcs.

Productivity of industry A = 30 1532 16

= and productivity of industry B =20 532 8

=

If the productivity is expressed in percentage, the same for A is 1516

× 100 = 93.75%

and productivity of industry B is 58

× 100 = 62.5 %

Production per week of industry A = 30 × l6=480 nos.

Production per week of industry B = 20 × 6= 120 nos.

Problem: 3 The following data is available for a machine in a manufacturing unit:

Hours worked per day 8

Working days per month 25

Number of operators 1

Standard minutes per unit of production

Machine time 22

Operator time 8

Total time per unit 30

(i) If plant is operated at 75% efficiency, and the operator is working at 100% efficiency, what is theoutput per month?

(ii) If machine productivity is increased by 10% over the existing level, what will be the output per month?

(iii) If operator efficiency is reduced by 20% over the existing level, what will be the output per month?

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Solution:

(a) Hours worked per day = 8Working days per month = 25Hours worked per month = 25 × 8 = 200 hrs.Machine time = 22 minutesOperator time = 8 minutesTotal per unit = 30 minutes = ½ hr.

No. of units produced/month/operator = 2001/ 2 = 400

As the no. of operator is 1, monthly production = 400 units. As the plant operates at 75% efficiency.

Monthly production = 400 × 75

100 = 300 units.

(b) If machine productivity is increased by 10% i.e. Machine time = 22 × 100

(100 10)+ = 20 minutes.

Then, total time = 20 + 8 = 28 minutes

Monthly production = 400 30 75

321units.28 100× × =

(c) If operator efficiency reduced by 20% i.e.

Operator time = 8 x (100 20)6

100+

= 8 × 7.2 = 57.6 minutes.

Total time = 22 + 57.6 = 79.6 minutes.

Monthly production = 400 30 75

113.06 units = 113 units. 79.6 100

× × =

Problem : 4

An incentive scheme allows proportionate production bonus beyond 100% performance level.

Calculate the amount of (i) incentive bonus and (ii) total payment received by an operator on a particularday during, which the following particulars apply:

Operation : Assembling pocket transistor radio setWork content : 30 standard minutes per assembled setAttended time : 8 hoursTime spent on unmeasured work : 2 hoursNumber of sets assembled during the day : 15Wage rate : Rs. 4/- per hour

(iii) What is the labour productivity achieved by the operator during the day?

Solution:

Attended time = 8 hrs. Time spent for unmeasured work = 2 hrs.

Actual time for production = 8 – 2 = 6 hrs.

Standard assembling period = 30 minutes/set. Standard production


177

= 630

× 60 set = 12 sets, in 8 hrs.

Wage rate = Rs.4/hr = 4 × 8/day = Rs. 32/day.

For 12 set of assembling Rs.32 is the wage

For 1 set of assembling 3212

is the wage

For 15 set of assembling = 3212

× 15 is the wage = Rs. 40/-

Total payment = Rs. 40/-Incentive bonus = Rs. (40 – 32)/- = Rs. 8/-

Labour productivity = S tan dard Time

Actual Time

Actual Time =6 60

2415×

= minutes.

Labour productivity = 30

100 125%24

× =

Problem: 5

The following data is available for a manufacturing unit :

No. of operators : 15

Daily working hours : 8No. of days per month : 25Std. production per month : 300 units

Std. Labour hours per unit : 8

The following information was obtained for November 2007:

Man days lost due to absenteeism : 30Unit produced : 240Idle Time : 276 man hours

Find the following:—

(a) Percent absenteeism

(b) Efficiency of utilisation of labour

(c) Productive efficiency of labour

(d) Overall productivity of labour in terms of units produced per man per month.

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Solution:

No. of days per month = 25

Daily working hrs. = 8

No. of operators = 15

No. of Man days = 15 × 25 = 375 Man days.

Total working hrs. = 375 × 8 = 3,000

Hours lost in absenteeism = 30 × 8 = 240

i) Percent absentees = 240 hrs. 100

8%3000 hrs

×=

ii) Efficiency of utilisation of labour = Standardlabourhourtoproduce240units 240 8

64%Totallabourhour 3000

×= =

iii) Standard time required to produce 240units = 240 × 8 = 1920 labour-hours.

In November, man hours lost = 30 × 8 = 240

„ „ idle time = 276

Total loss of time = 516 hours.

Productive hours available in November = 3000

Less, Total loss of time = 516

Actual labour-hours = 2484 hours

Efficiency of labour = Std.Labour hrs. 1920 100

77.3%Actual Labour hrs. 2484

×= =

v) 15 men produces 300 units,

Std. labour productivity = 300/15 = 20 units.

In November, overall productivity = 240/15=16 units. (Ans.)

i.e. productivity falls by 25%.

Problem: 6

A product is manufactured at the rate of 200 units per 4 day and sold for Rs. 8/- each. Direct material costis Rs.2/- per unit and direct labour cost is Re.1/- per unit, overheads (including selling) are Rs. 800/- perday. If the selling price can be reduced by Re.1/- per unit, it is expected that 50% more units can be sold.The workmen are prepared to produce 50% more only if there is a proportionate increase in their earnings.A suitable incentive scheme would cost Rs. 100/- per day to administer. With appropriate calculationsjustify if the Company should go in for such an incentive scheme.


179

Solution:

Production/day = 200 units/day = Rs. 200/- × 8 = Rs. 1600

Direct material cost = Rs. 2/-

Direct labour cost = Re. 1/-

Total = Rs. 3/-

Over head = Rs. 800/day = 800/100= Rs. 8/- per unit

Profit = 8 – 7 = Re. 1/per unit = Rs. 200/ day

If selling price = Rs. (8 – 1)/- = Rs. 7/- per unit

No. of unit sold / day = 200 × 150100

= 300

Turnover/day = 7 × 300 = Rs. 2100/-

Profit = Rs. (2100 - 1800 )/- = Rs .300/- per day.

So, it is justified for the company to go in for this incentive scheme.

Incentive either on individual or on group basis plays the following role in increasing productivity: –

i) Efficiency of production is increased.

ii) Labour cost and consequently cost of production can be reduced.

iii) Incentives ensure more earning of the labour force.

iv) It increases volume of production and makes the products available at reasonable price.

v) Incentive improves productivity and helps growth of national wealth at a faster rate.

Though incentive is very much beneficial to everyone related to it (employer, employee and indirectlythe society) but an incentive scheme tends to increase the material wastage if not properly guardedagainst.

Problem: 7

The targeted weekly output of a manufacturing unit employing 20 workers is 400 pieces. The group isentitled to earn an incentive @ 10% on the aggregate of wages based on basic piece rate plus dearnessallowance (which is Rs. 120 per week) upon achievement of a minimum of 80% of the output target.

This incentive rate increases by 21/2% flat for every 10% of increase in achievement of targets up to amaximum of 10% at the level of 120% of the output target in the following manner:

50 % increase in labour cost, i.e. it will be Rs. 300/- day.

Direct Material cost = Rs. 2/- unit Rs. 600/- day.

Overhead Rs. 800/- day.

Extra cost Rs. 100/- day.

Total cost Rs. 1800/- day.

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During the four weeks in February, the actual outputs achieved by the workers are 383 pieces, 442pieces, 350 pieces and 318 pieces respectively. The average basic piece rate is Rs. 5. Compute theamount of incentive earned by the group during each of the four weeks.

Solution:

Computation of Incentives in February

Problem: 8

Payment of bonus is made in a company on the following scale on the basis of the percentage of time savedon time allowed.

Time saved Bonus(% of standard) (% of time saved)

(a) Up to 25% 10%

(b) 26% to 30% 20%

(c) 30% and above 30%

You are required to calculate the earnings of A who takes 50 hours; B who takes 70 hours; C who takes 90hours. The standard time is 100 hours and the wage rate is Rs. 3 per hour.

Output target Incentive rate

80% - 90% 10%

90% - 100% 12 1/2%

100% - 110% 15%

110% - 120% 171/2%

120% and above 20%

Week1st 2nd 3rd 4th

Actual output achieved 383 442 350 318Achievement percentage(Actual output/ Targeted output) 95.75 110.5 87.5 79.5Wages at basic piece rate of Rs. 5 1915 2210 1750 1590Dearness allowance 120 120 120 120Total 2035 2330 1870 1710Incentive (%) 12.5 17.5 10 NilIncentive earned by group

(Total x incentive %) 254.38 407.75 187 nil


181

Problem: 9

Workmen of a particular grade working on 8-hour shift duty are guaranteed a wage of Rs. 32. An incentivescheme is in operation according to which production bonus is earned directly proportional to performancebut only after 100% performance is reached. Four workmen A, B, C and D produce 48, 60, 75 and 90 unitsrespectively in 6 hours spent on working on a job which has a standard time of 6 minutes per unit asmeasured work content. Remaining 2 hours of the shift are spent in doing unmeasured work for which noincentive bonus can be paid. Find for each workman: (i) the production performance level achieved,(ii) total earnings for the day.

Solution:

Rs. 32 Here the general base wage rate = Rs. 32 / 8 = Rs. 4 per hour

(i) Production performance levels

Workman A B C D

Standard units produced S = 48 × 6 60 × 6 75 × 6 90 × 6at 6 minutes per unit (s) = 288 = 360 = 450 = 540 mins.

Time worked on 360 360 360 360 mins.measured work for 6hours (T)

Thus, performance (288/360) × 100 (360/360) × (450/360)× (540/360) ×(S/T) x 100 = 80% 100 = 100% 100 = 125% 100 = 150%

(ii) On the basis of this information, let us calculate total earnings for the day.

A B C DEarnings on unmeasured work for 2 hours @ Rs. 8 Rs.8 Rs. 8 Rs. 8Rs. 4 per hour

Earnings on measured work for 6 hours @ Rs. 24 Rs. 24 Rs. 30 Rs. 36Rs. 4 per hour plus performance bonus whenperformance exceed 100%

Total earnings for the day Rs. 32 Rs. 32 Rs. 38 Rs. 44

A B CStandard time 100hrs. 100hrs. 100hrs.Time taken 50hrs. 70hrs. 90hrs.Time saved 50hrs. 30hrs. 10hrs.% of the time saved on standard time 50 % 30 % 10 %Time wages (Time taken x Rs.3) Rs. 150.00 Rs.210.00 Rs.270.00Bonus up to 25 % 7.50 7.50 3.00Bonus up to 30 % 3.00 3.00 -Bonus up to 30 % 18.00 - -Total earnings 178.50 220.50 273.00

Solution:

182


Here C and D get bonus of Rs. 6 and Rs. 12 respectively because their performance is 25% and 50%respectively higher than 100%.

Problem: 10

Fair Play Co. Ltd. has introduced a Scanlon Plan of Incentive Bonus for its employees in 200 based on thefollowing information relating to the previous three years:

Years Sales Revenue Total Salaries and Wages

2004 1,20,000 36,000

2005 1,25,000 35,000

2006 1,35,000 35,100

For 2007 the Sales Revenue has been Rs.1,50,000 and the total salary and wage payment has beenRs. 36,000. What is the amount due as Bonus to the employees according to Scanlon Plan? If 30% is set asidein a bonus equalisation reserve fund, how much money is available to be paid out as Scanlon Bonus for 2007?

Solution:

Ratio of Labour cost to sales revenue = Total Salaries & Wages / Sales Revenue

For, 2004 = 36,000/120,000 = 0.30 ; For 2005 = 35,000/125,000 = 0.28 and 2006

= 35,100/1,35,000 = 0.26

Average for three years = 0.84/3 = 0.28; Expected Labour Cost on Sale Revenue of 2007

= 0.28 × 1,50,000 = Rs. 42,000;

Actual Labour Cost (Salaries & Wages for 2007) = Rs. 36,000. Savings in labour cost due to increasedproductivity = Rs. 6,000. If 30% is set aside in a bonus equalisation reserve fund, the bonus for 2007will be = Rs. 6,000 – Rs. 1,800 (30% of 6,000) = Rs. 4,200.

Problem: 11

An incentive scheme allows proportionate production bonus beyond 100% performance level. Calculatethe amount of (i) Incentive bonus and (ii) Total payment received by an operator on a particular dayduring which the following particulars apply:

Operation : Assembling pocket transistor radio set

Work Content : 30 Standard minutes per assembled set

Attended Time : 8 Hours

Time spent on unmeasured work : 2 Hours

Numbers of sets assembled during the day : 15

Wage rate : Rs. 4 per hour

(iii) What is the net labour productivity achieved by the operator during the day?


183

Solution:

Total standard minutes worked during the day = 30 × 15 = 450, working time = 8 – 2 = 6 hours = 360minutes. Performance = (450 × 100) /360 = 125% or 0.25

(i) Incentive bonus = 0.25 × 6 × 4 = Rs. 6 for six hours on measured work

(ii) Guaranteed wage for 8 hours = 8 × 4 = Rs. 32; Total earnings for the days

= Rs. (6 + 32) = Rs. 38

(iii) Net labour productivity = Output in units / Net man hours = 15 / 6 = 2.5 sets per hour

Problem: 12

The following data is available for a machine in a manufacturing unit:

Hours worked per day 8

Working days per month 25

Number of operator 1

Standard minutes per unit of production: Machine time 22

Operator time 8

Total per unit 30


(ii) If machine productivity is increased by 20% over the existing level, what will be the output per month?

(iii) If operator efficiency is reduced by 20% over the existing level, what will be the output per month?

Solution:

(i) If plant is operated at 75% efficiency, output per month = (Actual Time / Standard Time) ×

efficiency level = 25 8 60 75

30 100× × × = 300 units. (ii) If the machine productivity is increased by

10% actual machine time becomes (22 × 100) / 120 = 18.33 minutes; Actual total time per unit =18.33 + 8 = 26.33 minutes.

(ii) Output per month = (8 × 60 × 25 × 75) / (26.33 × 100) = 342 units.

(iii) If operator’s efficiency is reduced by 20% operator time becomes = (8 × 100) / 80 = 10 mts; Actualtotal time per unit = 10 + 22 = 32 mts.; Output per month = (8 × 60 × 25 × 75) / (32 × 100) = 281.

Problem: 13

An operator is paid a flat rate of Rs. 30 per day and Re. 1 per every additional 5 points of efficiency ratingover a rating of 100 which represents standard performance, as incentive bonus. The standard minutes perunit of production are 20 and the operator makes 27 units in a shift of 7.5 working hours. (i) How muchincentive bonus he would get per day? (ii) If the incentive system is modified to pay the incentive at therate of 50% of the time saved, what incentive would he get?

184


Solution:

(i) Standard performance (100 points) = (7½ hours × 60) ÷ 20 = 22.5 units; Actual performance = 27units. Efficiency rating % = (27 / 22.5) × 100 = 120%; Incentive bonus per day = (20/5) × 1 = Rs. 4/-

(ii) Time required for producing 27 units; 27 × 20 = 540 minutes.

Time actually spent = 7½ × 60 = 450 minutes

Time saved = 90 minutes; Incentive bonus = 90 30

50%450× × =Rs. 3/-

Problem: 14

The following information about a company is available

Year Sales Revenue (Rs.) Total employeeRemmeration (Rs.)

2004 2, 40,000 72, 000

2005 2, 50, 000 70, 000

2006 2, 70,000 70, 200

For 2007, the sales revenue has been Rs. 3, 00,000 and the employee remuneration has been Rs. 72,000.What would be the amount due to the employees, if a Scanlon Plan is introduced?

Solution:

Average sales revenue for three years period= Rs. (2,40,000 + 250000 + 270000) / 3 = Rs. 253333.33/-Average employee remuneration = (72,000+70,000+70,200) / 3 = Rs. 70,733.33/-

Total bonus = Rs. 3,00,000 Rs.70,733.33

Rs.84,000 (approx.)Rs.2, 53, 333.33

× =

... Amount due to employees = 84,000 – 72,000 = Rs. 12,000/-

Problem: 15

Manufacture of a component requires operations to be performed on three machines P, Q and R respectively,the standard times and operator efficiency being as follows:

If the factory operates 2 shifts of 8 hours each and the machines are available for production throughout

Machine Standard hours per component Operator efficiency

P 0.16 80%

Q 0.23 100%

R 0.09 90%

the shifts on six days in a week, how many of machines P, Q and R will be required to produce 4,800components per week? How many hours of capacity, if any, would be available from the machines P, Qand R for doing other jobbing work?

Solution:

Available hours = 6 days × 2 shifts × 8 hours = 96 hours.Actual hours Required= (Standard hrs. per unit × Production quantity) / Operator efficiency


185

Machine P = (0.16 × 4800) / 80% = 960 hours; Machine Q = (0.23 × 4800) / 100% = 1104 hours

Machine R = (0.09 × 4800) / 90% = 480 hours; No. of machines required = (Actual hours required /Available hours); P = 960 / 96 = 10 machines; Q = 1104 / 96 = 11.5 i.e., 12 machines; R= 480 / 96 = 5machines.

Machines P and R will be fully utilised with no spare capacity. Machine Q will have ½ machine tospare, i.e., ½ × 96 = 48 hours per week can be utilised for doing other jobbing work.

Problem: 16

Workmen of a certain grade are guaranteed a wage of Rs. 48 for an eight hour shift. An incentive scheme isin operation which pays production hours directly proportional to production performance only after100% level is reached. Four workmen P, Q, R and S produce 90, 75. 60 and 48 units respectively in 6 hours.Work content of the job is 6 standard minutes per unit. Remaining 2 hours of the shift are spent on doingmiscellaneous unmeasured work for which no incentive hours can be paid.

Solution:

Performance of R at 100% and S at 80%. Both earn Rs. 48 for the 8 hours shift. This is a problem withincentive schemes with ‘ guaranteed base wage’ . If S does not improve to 100% performance after a period,he may be transferred to unmeasured work.

Problem: 17

An operator manufactures 11 identical components in a week of 48 hours duration. Each component takes 360standard minutes. Estimate the cost per component if the company operates an incentive system as below.

Guaranteed basic rate is Rs. 4 per hour upto 80% performance level, 11% of the basic rate is paid if theperformance level is between 80% and 100% and 120% of the rate is paid if the performance level fallsbetween 101 and 110%. Above 110% performance level, 130% of the basic wages are paid. Materialscomponent of each job is 150% of the direct labour.

Solution

Standard performance per week = (48 × 60) / 360 = 8 components

Actual performance per week = 11 components; Efficiency level = 11/8 × 100 = 137.5%. As the

Workmen P Q R S Units produced in 6 hours spent on measured work

90 75 60 48

Standard minutes per unit (s) 6 6 6 6 540 450 360 288 Minutes worked in measured work 6 hours (A)

360 360 360 360

Performance = S /A × 100 540

100360

× 450

100360

× 360

100360

× 288

100360

×

= 150% = 125% = 100% = 80% Earnings on measured work for 6 hours

= 36 × 1.5 = 36×1.25 = 36 × 1 = 36 × 1

= Rs. 54 = Rs. 45 = Rs. 36 = Rs. 36 Earning on unmeasured work for 2 hours @ Rs. 6 per hour

12 12 12 12

Total earning for the 6 hours shift = Rs. 66 = Rs. 57 = Rs. 48 = Rs. 48

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performance level exceeds 110%, 130% of the basic wages are payable.Wage Rate = (Rs. 4 × 130)/ 100 = Rs. 5.20 per hour

Labour cost (48 × 4 × 130)/(11×100) = Rs. 22.69

Material cost = Rs. 20.00

Over head cost @ 150% of direct Labour = Rs. 34.04

Total cost per component = Rs. 76.73

Problem: 18

A company manufactures 200 units of a product everyday and sells it for Rs. 8 each. Direct material cost isRs. 2 per unit and direct labour is paid Re. 1 per unit. Overheads are Rs. 800 per day in total. A marketresearch survey indicates that 300 units can be sold per day if the price can be brought down to Rs. 7.Production can be increased to this level, if an incentive scheme which would cost Rs. 100 per day toadminister is implemented, giving the workmen proportionate increase in their earnings. Examine withappropriate calculations whether and, if so, how much gains are made by the company and the workmen.Comment on the social desirability of such incentive schemes.

Solution:

Direct material cost per unit = Rs. 2; Direct labour cost per unit = Re. 1

Total revenue for 200 units = 200 × 8 = Rs.1600;

Total cost for 200 units = Rs. 800 + (200 ×3) = Rs. 1400; Profit: 200

Total revenue for 300 units = 300 × 7 = 2100;

Total cost for 300 units = Rs. 800 + Rs. 100 + (300 × 3) = Rs.1800

Profit: 300. Gains made by the company from the incentive scheme

= Rs. 300 – Rs. 200 = Rs. 100

Gains made by workmen = Rs.100. Such incentive schemes are socially desirable as both workmenand the company stand to gain. Consumers also gain because the product is available at a lower price.

Problem: 19

A company operates a wage incentive plan as follows:

Productivity Level Total incentive in a monthRs. (lacs)

Less than 75% Nil75% to 76% 176.1 to 77% 277.1 to 78% 378.1 to 79% 479.1 to 80% 580.1 and above 6

The total incentive is shared by the workers with hours worked by each as the basis. In a month the outputwas 80,000 standard hours. There are 600 workers, the total hours worked being 90,000 in the month,(i) What is the overall productivity? (ii) What is the total incentive? (iii) A worked for 150 hours and B for200 hours. What are their individual earnings?


187

Solution:

(i) Overall productivity= (80,000 / 90,000) × 100 = 88.89% (ii) Total incentive = Rs. 6 lacs (iii) Earningsof A = (6,00,000 /90,000) × 150 = Rs. 1,000; Earnings of B = (6,00,000 /90,000) × 200 = Rs. 1,333.33

Problem: 20

A soap factory adopts the piece rate system for its packing section, the rate being 10 paise. There is aguaranteed wage of Rs. 20 per day. The following data is available in regard to the number of soap cakespacked per day.

(a) (i) What is the wage payable to each worker? (ii) What is the average cost of packing per soap cakefor the day?

(b) The standard time for packing is 4 minutes in the above example.(i) What is the labour productivity ofeach worker for a shift duration of 480 minutes?(ii) What is the productivity of the group? (iii) Giveyour comments on the productivity figures obtained.

Solution:

(i) Wages payable are – (A) 800 × 0.10 = Rs. 80; (B) 600 × 0.10 =Rs.60; (C) 100 × 0.10 = Rs. 10

= Rs. 20 (guaranteed wage) ; (D) 700 × 0.l0 = Rs.70

(ii) Average cost of packing per soap for the day

= 80 60 20 70

100800 600 100 700

+ + + ×+ + + = 23000/2200 = 10.45 Paise approx.

(b) (i) labour productivity of each worker (A) : [800 ÷ (480/4)] × 100 = 666.7%; (B) : [600 ÷ 480 / 4]× 100 = 500%; (C): [100 ÷ 480 / 4] × 100 = 83.3% ; (D): [700 ÷ 480 / 4] × 100 = 583.3%

(ii) Productivity of the group = [(800 + 600 + 100 + 700)] ÷ 480 4

4×

× 100 = 458.3%,

(iii) Productivity figures of all the workers except C are excellent. Group productivity is also quitegood.

Problem: 21

The time study section of a tyre factory has fixed the following work standards for the tyre trimmingsection:

Worker No. of soap cakes packedA 800B 600C 100D 700

188


Production data for 2 groups are as follows —

(i) What is the productivity of each group? (ii) What is the overall productivity of both groupscombined? (iii) Each worker is paid a basic wage of Rs. 50 per day, D.A. of Rs. 60 per day andincentives as per the scheme shown below:

Size of Type Standard Time (Man – Mins)Tractor 10Truck 6Passenger-car 4Scooter 2

Group I Group IINo. of men 10 10Hours worked 8 8Production - Tractor 100 50Truck 200 250Passenger-car – 50Scooter 200 100

Productivity Rate / day (Rs.)

<49% 0

50 – 60 % 10

60 – 70 % 20

> 70 % 40

What will be the total wages of a worker in Group I and a worker in Group II?

Solution:

(i) Productivity of Group I:—

Standard time = 100 × 10 + 200 × 6 + 200 × 2 = 2600 Man-Mins.

Actual time taken = 10 × 8 × 60 = 4800 Man-Mins.

Productivity = (2600 / 4800) × 100 = 54%.

Productivity of group II:—

Standard time = 50 × 10 + 250 × 6 + 50 × 4 + 100 × 2 = 2400 Man-Mins.

Actual time taken = 10 × 8 × 60 = 4800 Man-Mins;

Productivity = (2400 /4800) × 100 = 50 percent

(ii) Overall productivity of both groups combined, standard time

= 2600 + 2400 Man-Mins = 5000 Man-Mins.

Actual time = 4800 + 4800 = 9600 Man-Mins;

Productivity = (5000 /9600) × 100 = 52 per cent

(iii) Total wages of a worker = Basic wage + DA + incentive

Group I = Rs. 50 + 60 + 10 = Rs. 120, Group B = Rs. 50 + 60 + 10 = Rs. 120


189

Problem: 22

A shop undertaking piece work and word processing job has the following working systems. There arefour operators A, B, C and D. Each one is given a minimum guaranteed wage of Rs. 50 for a working dayfor keying in and taking out printing of 25 pages. For any extra page they are given Rs. 3. Following datais available by each operator for a week of 6 days.

Operator Number of Excess over Incentivepages completed Standard output payment

A 200 50 50 ×3 = Rs. 150B 180 30 30 ×3 = Rs. 90C 150 0 0D 220 70 70 ×3 = Rs. 210

Operator Number of pages completedA 200B 180C 150D 220

(a) What is the remuneration payable to each worker? Does the cost per page depend on quantity?Any variation from normal? (b) How do you incorporate quality incentive in the system?

Solution:

(a) Standard production = 25 × 6 = 150 pages. Standard remuneration = 50 × 6 = Rs. 300

Thus, the remuneration payable to each operator is as follows:

A Rs. 300 + Rs. 150 = Rs. 450

B Rs. 300 + Rs. 90 = Rs. 390

C Rs. 300 + 0 = Rs. 300

D Rs. 300 + Rs. 210 = Rs. 510

Cost per page does not depend upon quantity. The incentive scheme is unbalanced due to low benefitfor management.

Problem: 23

Workmen in a particular grade are guaranteed a daily wage of Rs. 64 for an eight hour shift. According toan incentive scheme in operation, incentive bonus is earned directly proportional to production above100%. Four workmen A, B, C and D produce 48, 60, 75 and 90 units respectively in six hours spent inworking on a job which has standard time of 6 minutes per unit. The remaining 2 hours are spent in doingsome unmeasured work for which no incentive bonus can be paid. For each of the four workmen find:

(i) the level of production performance achieved ;(ii) the total earnings for the day.

Solution:

Here the general base wage rate 64/8 =Rs. 8 per hour

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(i) The level of production performance achieved:

Worker Nos. Piece rate (Rs.) Guaranteed rate Earnings payablepacked (Rs.)

A 400 80 35 80

B 300 60 35 60

C 175 35 35 35

D 150 30 35 35

1025 210

Worker No. of cakes packedA 400B 300C 175D 150

Workmen A B C DStandard units produced at 6 minutes per 48 × 6 60 × 6 75 × 6 90 × 6unit (S) = 288 = 360 = 450 = 540Time worked on measured for 6 hours (T) 360 360 360 360Performance (S/T) x 100 80 100 125 150

Earnings on unmeasured work for 2 hours @ Rs. 8 per hour 16 16 16 16 Earnings on measured work for 6 hours @ 48 48 48 48 Rs. 8 per hour Performance – – 12 24

Rs. 64 64 76 88

(ii) Total earnings per day:

Problem: 24

The packing section in a detergent factory operates under the guaranteed piece rate system of wage payment.The piece rate per cake packed is 20 paisa. The guaranteed wage per day is Rs. 35. On a typical day, theproduction of four workers in the packing department is as follows:

(i) What are the earnings of each worker on this day?

(ii) What is the average labour cost of packing on this day?

(iii) The management wants to do away with the present individual incentive system and instead introducea group incentive system for all the 4 workers together (i.e.) giving them a flat incentive of Rs. 15 perday per worker if the combined production for the day exceeds 1,000 nos. and Rs. 20 per day perworker if the output exceeds 1,300 nos. What would be your advice to the management in this regard?You may assume a standard time per unit as 1.333 minutes and that the management saves Rs. 5 perday in the administration of the new group incentive scheme as compared to the earlier one.

Solution:

Given piece rate = Re. 0.20 per piece. Guaranteed wage rate = Rs. 35 per day. Payment as per guaranteedpiece rate system.

(i) Computation of earnings:


191

Job Cards Std. Hrs No. of persons1 17 32 15 23 19 34 18 35 23 56 16 57 22 38 14 4

(ii) Average labour cost or packing: Labour cost = Rs. 210; Nos. packed = 1,025; Cost per pack = Re. 0.20 (approx.) (iii) Evaluation of the group incentive scheme: Standard time per unit = 1.333mins.; Standard output for 4 persons = (480/1.33) × 4 = 1,440; Earnings based on this scheme inthe present case: Normal wage = Rs. 140;

Incentive = Rs. 15 × 4 = Rs. 60; Actual cost = Normal wage + Incentive – Savings.

= Rs. (140 + 60 – 5) = Rs. 195

Since the group incentive costs are less than the present guaranteed scheme, management should gofor it.

Problem: 25

(a) Find out the productivity index of a section from the following figures:

(i) Standard Man-Hrs:

Reasons Idle time (Hrs.) No. of persons involvedNo work 10 1No Material 5 2Power Failure 10 15No Material Handling Equipment 2 3Machine Breakdown 10 4Others 5 3

Workers Attended Hrs.3 1805 1704 1603 150

(ii) Idle Card:

(iii) Record of attended hours for direct workers :

Compute: Productivity Index and Wastivity Index.

b) Do you agree with the view “Job security is an impediment to productivity”? Justify your answer.

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Solution:

(a) Standard Man Hrs Produced (S.M.H.) would be = Std. hrs x no. of persons for each job.

= 17 × 3 +15 × 2+ 19 × 3 + 18 × 3 + 23 × 5 + 16 × 5 + 22 ×3 + 14 × 4 = 51 + 30 + 57 + 54 + 115 + 80 +66 + 56 = 509 hrs.

Attended Man-hrs (A.M.H.) = Attended hrs. × No. of workers

(as per record of attended hrs. for direct workers — as given)

= 180 × 3 + 170 × 5 + 4 × 160 + 3 × 150

= 540 + 850 + 640 + 450 = 2480 hrs.

Idle man-hrs = Idle Hrs (as given) × no. of persons involved = 10 × 1 + 5 × 2 +10 × 15+ 2 × 3 +10 × 4 +5×3 = 231 hrs.

Effective A.M.H. = 2480 – 231 = 2249 hrs. Productivity index (P.I.) = (509 / 2,249) × 100 = 22.63%

Wastivity = 100 – 22.63% = 77.37%

(b) I agree with this statement especially with reference to our country. In general, a worker has noincentive or pressure to improve his productivity when there is no likelihood of losing his jobdespite low levels of performance. In Government jobs, in public sector and in industries wherepowerful trade unions exist, productivity tends to be low because workers know them. In thistype of work culture it is not possible to increase productivity. Therefore, general productivitylevels remain low. On the other hand, if workers know that they will lose their jobs in case theyfail to give a minimum level of productivity; they would work hard to retain their jobs. Therewould be a continuous pressure on them to perform better. According to behavioural scientistshuman beings tend to perform better under a little tension. This tension at the work place can becreated if job security is not guaranteed. At the same time safeguards must be provided to ensurethat a worker is not thrown out of the job on frivolous grounds or for reasons which are not hisresponsibility. Suitable legislation can be created to protect workers against victimisation. Thus,absolute job security is an impediment to productivity in India.

Problem: 26

A product is manufactured at the rate of 200 units per day and sold for Rs. 8 each. Direct Material cost isRs. per unit and direct labour cost is Re. 1/- per unit, overheads (including selling) are Rs. 800 per day.

If the selling price can be reduced by Re. 1/- per unit, it is expected that 50% more units can be sold. Theworkmen are prepared to produce 50% more only if there is a proportionate increase in their earnings. Asuitable incentive scheme would cost Rs. 100 per day to administer. With appropriate calculations, justifyif the company should go in for such an Incentive Scheme.


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Solution:Comparative cost and profitability: (per day)

Particulars Without WithIncentive (Rs.) Incentive (Rs.)

Units produced & sold 200 300Selling price – Rs. 8 7

Sales value – Rs. 1,600 2.100

Direct cost: Direct Materials at Rs. 2 400 600per unitDirect Labour cost atRs.1/- per unit 200 300

600 900

Contribution : (Sales - Direct cost) 1,000 1,200Overheads 800 800

Incentive __ 100Profit 200 300

Daily working hours 8

No. of working days in a week 6

No. of operators 20

Std. Hours per unit of production 4

Number of units produced 48

Absentee man days 40

Idle time due to load shedding 30 Mandays

Problem: 27

The following information is available for a factory:

During a particular week

Find:

(i) Absenteeism percentage

(ii) Labour utilisation percentage

(iii) Productive efficiency of labour

(iv) Overall productivity of labour in terms of units produced/week/employee.

Solution:

(i) Absenteeism percentage = Absentee mandays / week 40

100 33.33%Total operator mandays/week 2 6

= × =×

(ii) Labour utilisation percentage = Working mandays/ week

100Attendedmandays/weeks

×

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Total operator mandays/week Absenteeism Idle time100

Attended mandays− −

= ×

=(20 6) 40 30

100 62.50%(20 5) 40

× − − × =× −

(iii) Productive efficiency of labour = Std.man hours produced/ week

100Actual man hours worked/week

×

4 48100 48%

[(20 6) 40 30] 8× × =

× − − ×

(iv) Overall productivity of labour = Total production/week

48 / 20Total operators

=

= 2.4 Units/Week /operatorProblem: 28

A factory can manufacture two products A and B by using either of two materials P or Q. A is expected tosell at Rs. 70 per unit and product B at Rs. 30 per unit.

Material P Material Q Output A 200 units 400 units B 300 units 200 units Quantity, of raw material usage 1,000 kg 1,000 Kg Labour usage 300 man hrs. 250 man hrs. Electric energy consumption 1000 KWhr 1500 KWhr Cost of raw material/kg Rs.20 Rs.30 Labour per manhour Rs.5 Rs.5 Electrical energy/KWhr Rs.1.5 Rs.1.5

The operating data are as follows:

Compare the productivity of material, labour and electrical energy in using materials P and Q. Commenton the relative advantage of using either of the materials.

Solution:

Productivity = Value of outputValue of input

Sales value of output with material P

= Output of product A in units × Rate/unit of A

+ Output of product B × Rate/unit of B= 200 × 70 + 300 × 30 = Rs. 23,000

Sales value of output with material Q = 400 × 70 + 200 × 30 = Rs. 34,000

The partial productivity of different factors of production are computed as follows:


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Comments:

The productivities of (1) and (3) is nearly same by using either material P or Q. If labour is the key factor itis better to use material Q as labour productivity, for Q is higher, i.e., 27.200 > 15.333

Problem: 29

The following data is available for a machine in a manufacturing unit.

Material P Material Q 1. Productivity of raw materials

= used material raw of Value

output of valueSales

201000

23000

×= 1.150

30000

34000 =1.133

2. Labour productivity, i.e.,

= labour of Value


5300

23000

×=15.333

5250

34000

× =27.200

3. Electrical energy productivity, i.e.,

= energy electrical of Value


5.11000

23000

× =15.333

5.11500

34000

×=15.111

Number of hours worked per day 8 Working days per month 25 Number of operators 1

Machine time 22 min Operator time 08 min Total time/unit 30 min

Standard time per unit of production,


(ii) If the machine productivity is increased by 10% over the existing level, what will be the output permonth?

(iii) If the operator efficiency is reduced by 10% over the existing level, what will be the output per month?

Solution:

(i) Plant is operated at 75% efficiency

Machine time = 22 × 10075

= 22 ×43

min

Operator time = 8 min

Total time/unit = 22 4 112

83 3× + = min

Total time available = 8 × 60× 25 min

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Output Target Incentive Rate

80% - 90% 10%

90% - 100% 121/2 %

100% - 110% 15%

110% - 120% 171/2 %

120% and above 20%

Number of units produced per month 8 60 25(112 / 3)× ×

= 321.43 ≈ 322 units.

(ii) If the machine productivity is increased by 10%, i.e., plant efficiency = 75 + 10 = 85%.

Machine time = 22

0.85= 25.88 min

Operator time = 8 min (assuming 100% efficiency)

Total time/unit = 33.88 min.

Number of units produced/month = 8 60 25

33.88× ×

= 354.19 ≈ 355 units.

(iii) If the operator efficiency is reduced by 20%, plant efficiency is 75%.

Machine time = 22

0.75 = 29.33 min

Operator time = 8

0.9= 9 min/unit (operator efficiency 90%)

Total time/unit = 29.33 + 9 = 38.33 min

Number of units produced/month = 8 60 25

38.33× ×

= 313 unit S

Problem: 30

The targeted weekly output of a manufacturing unit employing 20 workers is 400 pieces. The group isentitled to earn an incentive @ 10% on the aggregate of wages based on basic piece rate plus dearnessallowance (which is Rs. 120 per week) upon achievement of a minimum of 80% of the output target. Thisincentive rate increases by 2 1/2% flat for every 10% increase in achievement of targets upto a maximum of10% at the level of 120% of the output target in the following manner :

During the four weeks in February, the actual output achieved by the workers is 383 pieces, 442 pieces, 350pieces and 318 pieces respectively. The average basic piece rate is Rs. 5. Compute the amount of incentiveearned by the group during each of the four weeks.


197

Solution:Calculation of incentives in February

2.13 Quality Control

We are all living in a competition age. In order to combat competition, every manufacturer would like tomaintain the quality of his products so that it may be liked by the users in preference to products of hiscompetitors. Reputation for uniformity and dependability of the quality of a product is one of the mostimportant assets which a manufacturing concern should try to acquire or not to lose if already acquired.

There is an old saying heard very often in business circles to the effect that one is reminded of quality longafter price is forgotten. Yet the management often fails to appreciate fully the truth of this statement. At thetime when production schedules are pressing, the tendency quite often is to sacrifice quality in favour ofquantity. The tendency is reflected in customer ill-will and reduced sales volume. Poor quality or reducedquality costs the concern much indirectly.

So, in order to maintain the quality of the product, quality control is applied at the work place. Quality ofraw materials and the parts and the machines and equipment should be inspected quite strictly becausequality of product depends much upon the quality of raw materials. During the process of production,several checks are done through inspection of the semi-finished product and finished product and it isconfirmed whether the quality of the product conforms to the standards and specifications already set. Forthis purpose, several devices or techniques are applied. If the quality of the product, it is found duringquality control process, is found not conforming to the specifications, the steps should be taken by theproduction department to improve the quality of the defected articles by reworking on those articles and ifit, anyhow, is not possible, other measures to sell those sub-standard items should be taken such as sellingsuch items at reduced rates as ‘B’ quality items.

This topic deals with the various devices or techniques of quality control (including statistical qualitycontrol and inspection).

Quality Control – Meaning, Objects and Importance

Meaning of quality control : The term ‘quality control’ consists of two words ‘quality’ and ‘control’. Qualityis that characteristic or a combination of characteristics that distinguishes one article from the other or

Week

1st 2nd 3rd 4th

Actual output achieved 383 442 350 318

Achievement percentage

(Actual output/ Standard output) × 95.75% 110.5% 87.5% 79.5%

100 (383/400) × 100

Wages at piece rate at Rs. 5/- Rs. 1,915 Rs. 2,210 Rs. 1,750 Rs. 1,590

Dearness allowance 120 120 120 120

2,035 2,330 1,870 1,710

Incentive % 121/2 % 17 1/2

% 10% Nil

Incentive earned by group Rs. 254.38 Rs. 470.75 Rs. 187 Nil

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goods of one manufacturer from that of competitors or one grade of product from another when both arethe outcome of the same factory. The main characteristics that determine the quality of an article mayinclude such elements as design, size, materials, chemical composition, mechanical functioning, electricalproperties, workmanship, finish and appearance. The quality of a product may be defined as the sum of anumber of related characteristics such as shape, dimension, composition, strength, workmanship,adjustment, finish and colour.

‘Control’ may be referred to as the comparison of the actual results (finished product) with the predeterminedstandards and specifications. It locates the deviations and tries to remove them. Control is the correction inthe quality of the produce when deviations in the quality are more than expected in the process. Controlconsists in verifying whether everything occurs in conformity with the plan adopted, the instructionsissued and principles established. It has for object to point out weaknesses and errors in order to rectifythem and prevent recurrence. It operates on everything – things, people and action.

Thus, by the term quality control, we mean the process of control where the management tries to conformthe quality of the product in accordance with the pre-determined standards and specifications. It is asystematic control of those variables that affect the excellence of the ultimate product.

Quality control may be defined as that industrial management technique or group of techniques by meansof which products of uniform acceptable quality are manufactured.

Quality control refers to the systematic control of those variables encountered in a manufacturing processwhich affect the excellence of the end product. Such variables result from the application of materials,men, machines and manufacturing conditions.

Thus quality control is a technique of scientific management which has the object of improving industrialefficiency by concentrating on better standards of quality and on controls to ensure that these standardsare always maintained. In this way, for quality control purposes, first standards and specifications areestablished and then to see whether the product conforms to those standards.

Objectives of quality control: The following are the main objectives of quality control programme:

1. To assess the quality of the raw materials, semi-finished goods and finished products at various stagesof production process.

2. To see whether the product conforms to the predetermined standards and specifications and whetherit satisfies the needs of the customers.

3. If the quality of the products deviates from the specifications, able to locate the reason for deviationsand to take necessary remedial steps so that the deviation should not be recurred.

4. To suggest suitable improvements in the quality or standard of goods produced without much increaseor no increase in the cost of production. New techniques in machines and methods may be applied forthis purpose.

5. To develop quality consciousness in the various sections of the manufacturing unit.

6. To assess the various techniques of quality control, methods and processes of production and suggestimprovement in them to be more effective.

7. To reduce the wastage of raw materials, men and machine during the process of production.

Importance or advantages of quality control system. The programme of quality control is advantageous toproducers and consumers both. A quality product will satisfy the customer’s needs on the one hand and


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consequently the demand of the product will increase resulting in large-scale production. On the otherhand, the goodwill of the firm increases as the producer of quality goods. It helps the producer in increasingthe market for the goods. The importance of quality control lies in the following facts:

• Reduction in costs: An efficient quality control system reduces the cost of production of the productdue to (i) reduction in wastage of raw materials, semi-finished and finished goods; (ii) large-scaleproduction of standard quality product; (iii) rework cost of the substandard goods is the minimum.

• Improvement in the morale of employees: By quality control programme, the employees becomequality-conscious. They understand the standards of the product well and try to improve the standardsand produce the quality goods to the best of their efforts. Thus it improves the morale of the employees.

• Maximum utilisation of resources: By establishing the quality control system, the necessary controlover the machines, equipment, men and materials and all other resources of the company is exercised.The system will also control the misuse of facilities, wastages of all types and low-standard production.Thus, the resources of the company are put to maximum use.

• Increase in sales: Increase in sales of the product is the main objective of the quality control system.By introducing quality control programme in manufacturing process, a quality product is madeavailable to the consumers and that is too at lower rates because of lower cost of production. It inturn increases the demand of company’s product.

• Consumer’s satisfaction: Consumers always get the quality products of standard specifications whichthey find to their utmost satisfaction.

• Study of variations: It is a well-known fact that some variations are bound to exist in the nature ofproduction inspite of careful planning. The magnitude of variations depends upon the productionprocess namely machines, materials, operations etc. The techniques of quality control helps in thestudy of these variations in quality of the product, and serves as a useful tool for the solution of manymanufacturing problems which cannot be solved so well by any other method.

Thus quality control is an important technique in the hands of management to maintain the quality of theproduct.

Meaning of inspection: Inspection is an important and essential tool of quality control that ascertains andcontrols the quality of a product. The main purpose of quality inspection is to safeguard quality by comparingmaterials, workmanship and pro ducts with the set standards. Inspection is a method by which the inspectormay decide how much of the total work done conforms with the pre-determined standards or how muchis below-standards so that a decision may be taken for its approval or rejection. If it conforms to specifications,it will be accepted otherwise the whole lot or a part of it which does not conform to standards will berejected. If defects in the items are beyond acceptable limits, a corrective action can also be suggested sothat the future production should be strictly according to the specification because, the fundamental purposeof inspection is to have an effective check on the production of defective items to ensure the quality of theproduct and to lower down the cost of production. Thus inspection can also be termed as a sorting processon the basis of which products can be classified into acceptable or unacceptable.

Inspection is the art of comparing materials, products or performances with established standards. Therecan be no intelligent inspection without definite standards. In any such items that are to be inspected,some will fall outside a liberal allowance of variation from the standards, some will be well within thelimits of error, and others will be very close to the limits. Inspection is the art of selecting these three classesof product which will be satisfactory for the work.

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Objectives of inspection: The main objectives or functions of inspection are as follows:

• Maintenance of quality: The fundamental purpose of inspection is to maintain the quality of theproduct. This function is performed by comparing materials, semi-finished or finished products, menand machines and tools with the established standards. Items which conform to the specifications orare within the acceptable limits are accepted and other items are rejected.

• Improving the product quality: By comparing the quality of the products against the set standards,the defective items are located and probable reasons for the defects are established. Necessaryadjustments are done for future by removing the reasons for defects and thus the quality of the productis improved steadily and regularly. It helps in safeguarding the prestige and confidence of theorganisation in the eyes of the consumers.

• Reduction in costs: As raw materials are inspected to see whether they are as per standards or not thedefective raw materials are thus not allowed to be used in production. Thus it saves the organisationfrom loss if any and reduces the costs of production.

Statistical Quality Control:

Inspection is an integral part of the scheme of production control. By inspection, quality of the product isconformed with the standards and specifications during or at the end of the production process. It isimperative to have certain disturbances which cause the product to deviate slightly from the desiredstandards. This type of variability is inherent in the process and is known as variability due to chancecauses. Causes of deviations like conditions of production process, nature of raw materials, the behaviourof operations etc. Such causes are assignable causes. Due to these causes, defective items are producedwhich lower the quality of the product.

Inspection has rather a limited use. It does not provide the extent to which certain products requiring ahigh degree of precision conform to the strict standards set in this regard. Similarly, it does not provide allthe information on the basis of which sub-standard finished products can be scrapped in the case of massproduction industries, producing standardised products. These imperfections of inspection in quality controlmake the use of statistical quality control indispensable in these cases.

Statistical quality control is the application of statistical techniques to -determine how far the productconforms to the standards of quality and precision and to what extent its quality deviates; from the standardquality. The purpose of statistical quality control is to discover and correct only those forces which areresponsible for variations outside the stable pattern. The standard quality is pre-determined through carefulresearch and investigation.

It is quite impracticable to adhere strictly to the standards of precision, especially in cases where humanfactor dominates over the machine factor. Some deviation is therefore, allowed or tolerated. They arereferred to as tolerances. Within the limits set by these tolerances, the product is considered to be of standardquality. SQC brings to light the deviations outside these limits.

Techniques of statistical quality control. The techniques of statistical quality control can be divided intotwo major parts:

1. Control charts, and 2. Acceptance sampling:

1. Control chart: Control chart is the most important quality control technique. It is a chart and depictsthree lines on the chart. One line is the central line showing the average size. The other two lines, one


201

below the central line and the other above the central line, indicate the limits of tolerances, withinwhich deviations from standards are permissible. The actual measurement of the whole lot or a sampleis plotted on the chart. Those measurement values which fall outside the tolerance limits are consideredto be out-of-control points and assignable cause may be said to exist. This will enable die manufacturersto know the causes of variation or causes of trouble which he can amend.

2. Acceptance sampling: It can be described as the post-mortem of the quality of the product thathasalready been produced. Under this technique, a sample is selected at random to examine whetherit conforms to the standards laid down. It can be assumed that a certain percentage of goods will notconform to the standards, so a certain percentage of defective products in a lot may be specified. Thistechnique has all the limitations of sampling technique. There are two limiting levels of quality in anacceptance sampling plan:

(i) the acceptable quality level (AQL) that represents the lowest percentage of defectives which abuyer is expected to accept and seller is expected to supply, and

(ii) the lot tolerance percentage defective (LPTD) that represents a limit at which the buyer wants tobe quite certain that the lot will not be passed.

The two limiting levels decide the risks to be borne. The AQL involves the producer’s risk or the risk thata lot with an acceptable quality will be rejected on the basis of sample inspection. The greater the risk of theproducer, the higher will be the charges for the product. On the other hand, risks attached to LPTD arecalled the consumer’s risk. Since only a sample is inspected under the plan, it is quite likely that an itemwhich does not conform to standards may be accepted on the presumption that the sample is typical of thewhole lot. The acceptance sampling inspection may be of any of the following types:

(i) Inspection by attributes,

(ii) Inspection by variable, and

(iii) Inspection by number of defects per unit.

Importance or benefits of statistical quality control: The technique of statistical quality control has becomevery popular since the days of World War II. In modem industry, it becomes a necessity since it offers thefollowing benefits:

• It saves on rejection: In the absence of statistical quality control technique, many products may befound defective and worthless at the end of manufacturing process and may be thrown away asscrap. SQC avoids such a situation and saves the cost of labour and material involved in the productionof defective items. It measures the extent of defect and defective products may be improved by re-working to the level of acceptable standards. It also saves the loss which will arise out of re-workingon the items rejected outrightly or which cannot be brought to the acceptable standard.

• It maintains, high standard of quality: The statistical quality control ensures the maintenance ofhigh standard of quality because lower standard of quality products are not put to the market. Theyare improved, if possible or rejected outrightly but in no case, sub-standard item is sold in the market.Thus, the concern gains in goodwill.

• Reduced expenses of inspection: It reduced the expenses of inspection to a great extent and enablesthe product to be manufactured at lower cost.

• Ensures standard price: If certain products are not up to the desired standard of quality and cannot

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be improved without much expense, they can be downgraded and sold cheaper. SQC maintains thestandard price for all standard products. Thus, it increases the profitability of the concern.

• Feelings of responsibility among workers: Among workers, a feeling of responsibility developsbecause they begin to understand that their work is being inspected very minutely hence they workcarefully. It helps increasing their moral.

Benefits of SQC:

(i) It saves on rejection

(ii) It maintains high standard of quality

(iii) It reduces expenses of inspection

(iv) It ensures standard price for all products

(v) Feelings of responsibility among workers development

Control Charts:

The concept of control charts and their application in production process was evolved for the first time in1924. This concept is based on the division of observations in rational sub-groups. The sub-groups areformed in such a way that the variation within each sub-group is attributable due to chance causes onlyand variation between the sub-groups are attributed due to assignable causes. The most obvious basis forthe selection of sub-group is the order of production. The problem of process control can be solved byapplying a method which helps in finding out the quality characteristics of various sub-groups. If thequality characteristics of various sub-groups are identical, the process is said to be in control otherwise outof control.

Statistical basis for control charts: Suppose Q is the quality characteristic of a product. This characteristicshould be measured for a number of sub-groups. If Q follows some standard distribution (Normal, Poissonor Binomial) with mean MQ and standard deviation óQ. If the sample size is sufficiently large then everydistribution will be normal and due to the assumption of normality the interval MQ± 3 σ Q will coverapproximately 99.73% values of Q. If all the observed values of Q lie within the limits MQ+3 σ Q, this maybe taken as the indication of absence of assignable cause. If any value of Q lies outside the limits, thepresence of assignable cause can be presumed and the process should be corrected before it is allowed tocontinue.

The control chart is a horizontal chart where time variable is taken on X-axis. A control line (CL) is drawnalong the X-axis at MQ point. Time for two other limits, lower control limits LCL (MQ-3 σ Q) and uppercontrol limit UCL (MQ+3 σ Q) are drawn taking the CL in the centre. The values of quality characteristicsQ are also plotted on group for different observations from sample to sample or according to time variables.

The following observations can be had from the control chart.

(i) If all the plotted points for Q lie within the control limits, the process is said to be in control otherwiseit can be said to be out of control with respect to the quality characteristic Q.

(ii) If the points on a control chart lie close to one of the control lines or the points show some specialtrend, then it is very difficult to pay anything about the process control. A typical control chart can bedrawn in the following figure—

X


203

Thus control chart is a graph showing the range of expected variability. The following points should beconsidered while drawing a control chart.

(i) If two comparable control charts are to be drawn on the same paper, it is better to take the same scalefor both the charts.

(ii) The charts should be narrow in comparison to their length.

(iii) So many charts should be avoided on the same sheet.

Types of Control Charts: Control charts are of two types: (i) control chart for variables, and (ii) controlchart for attributes.

If the quality characteristic Q is capable of quantitative measurement, the control chart is known as controlchart for variables and if the quality characteristic Q cannot be measured quantitatively the control chart isknown as control chart for attributes. The items can only be classified as defective or non-defective etc.

1. Control Chart for Variables. In this case, quality characteristic Q is capable of direct quantitativemeasurement and is a continuous variable. In such cases the average and the standard deviation ofthe quality characteristic can be calculated from a number of samples each having fixed number ofcomponents, taken at random over a period of time from any process. The following situations can beencountered in practice:

(i) The process may be in control.

(ii) The mean of the characteristic is out of control but standard deviation is not. For this purpose, controlchart for means (X-chart) are prepared.

(iii) The standard deviation is out of control and not the mean. This is studied by control chart for range(R-chart). Here range is the measure of variability because it can be easily determined.

(iv) Both mean and standard deviation are out of control. This is studied by X and R charts simultaneously.

(A) Control chart for mean ( X -chart):

By this chart, it is found whether the mean of the characteristic can be determined by the following formula:

Central line = =

=∑X

XK

UCL (MQ + 3 σ Q)

CL (MQ)

LCL (MQ – 3 σ Q)

0 1 2 3 4 5 6 7 8 9 10 11 XTIME VARIABLE OR SAMPLE NUMBER

FIG. CONTROL LIMITS

QU

ALI

TY

CH

AR

AT

ERIS

TIC

(Q

)Y

X

204


Where =X = Mean of the sample means

X∑ = Sum of the sample means

K = No. of samples.

By this method lower control limit (LCL) and upper control limit (UCL) are determined as follows:

(i) Where standard deviation is known –

UCL = 3= σ+ P

XN

+ and LCL = 3= σ− P

XN

where UCL = upper control limit

LCL = lower control limit

=X = mean of the sample means

Pσ = standard deviation of population

N = total number of items in a sample.

(ii) Where standard deviation is not given –

The standard deviation can be determined by the following formula:

2

Pσ = Rd

Where R = mean of sample range

d2 = quality control factor.

(iii) Where d2 is not given, then

UCL = =X + A2 R

LCL = =X - A2 R

where A2 = quality control factor.

(B) Control chart for range or range chart (R-chart). This chart is used to see whether the standard deviationof the characteristic Q is in control or not. Here can be two situations, viz, (i) where σ P is known. (ii)where σ P is unknown.

(i) Where σ P is known. In this case central line (CL) and upper and lower limits can be determinedas follows:Range (R) = Highest value – lowest value

Mean of R or R = ∑R

K

Central line = R = d2 Pσ where K = total no. of samples.

UCL= d2 Pσ + 3 Pσ .d3

LCL= d2 Pσ – 3 Pσ d3

Where d2 and d3 can be determined from statistical tables.


205

(ii) Where σ P is unknown

CL = R = R Sum of samplerange

K No.of Samples=∑

UCL = D4 RLCL = D3 R

Where D3 and D4 can be read from the statistical table for control charts. As a rule if LCL is negative, it istaken to be zero.Practical limitations of X and R-charts: X and R charts are very important tools for the diagnosis ofquality problems in a manufacturing process but they have the following practical limitations:1. These charts can be used where quality characteristic can be measured quantitatively.2. Quantitative measurement of quality characteristic is a costly proposition, hence its use is uneconomicalInterpretation of X and R-charts. X and R-chart can be used simultaneously to judge the quality of theprocess as follows:(i) Where R - chart shows all the points within control limits:

(a) If in X - chart, the points lie beyond one of the control limits, it shows that process level hasshifted.

(b) If in X - chart, the points lie beyond both of the control limits, it shows that the process level ischanging at random and needs frequent adjustment.

(ii) Where R-chart shows variability out of control:(a) If X - chart shows points beyond one of the control limits, it signifies that both process level and

variability have changed.(b) If X -chart shows points beyond both of the control limits, then this implies that variability has

increased.(iii) If points in X and R-charts are too close to the central line, then it shows that there exist systematic

difference within sub-groups.2. Control chart for attributes: Where the nature of product is such that the quality characteristic cannot

be measured quantitatively, the items are classified only defectives and non-defectives at the time offinal inspection.There can be a number of factors responsible for defining any item to be defective and the separate recordfor each cause maybe out of question. The defects can be measured in airy of the following two ways:(i) Number of defective items are taken in different samples. Here P or n P charts is used.

(ii) Number of defects in one item. In this case C chart is used.

Control chart for fraction defectives ( P or n P charts )

These charts are constructed by recording at least 20 successive inspections. The percentage of defectiveitems is then calculated. The limits for P-chart are given by:

CL or =×∑P

Pn k

UCL or upper control limit = P(1 P)

3n−+P

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LCL or Lower control limit = P(1 P)

3n−−P

Here P = central lineP = No. of defective items in a sampleK = Total no. of samplesn = Sample size (no. of items in a sample),

nP chart = In situations where the number of items inspected in each sample is the same, nP charts areprepared. The various control limits in this case are

CL = n ∑PP =

K

UCL n (1 )= −P + 3 P P

LCL n (1 )= −P - 3 nP P

P and nP charts can be used in the case of variables also.C-charts. C = charts are prepared where defective items are taken out by the number of defects in oneitem. Items which are according to specifications are termed as standard items. Items which have oneor more defects, it means they do not fulfils one or more of the given specifications. All defects are notof the same value. So, we may like to control defects per unit. The quality characteristic in such casesis the number of defects per unit. C-chart is just an improvement over P-chart. C-chart follows Poissondistribution and various limits are calculated as follows:

Central line = Totaldefects

TotalNo.of itemsinspected=C

Upper control limit (UCL) = C + 3C

Lower control limit (LCL) = C – 3C


Problem 1 :

Draw the control charts for X (mean) and R (Range) from the following data relating to 20 samples, eachof size 5. Only the control line and the upper and lower control limits may be drawn in each chart.

Sample No. X R Sample No. X R1 38.2 15 11 32.6 312 33.8 1 12 22.8 123 24.4 22 13 21.6 294 36.6 24 14 28.8 225 27.4 18 15 28.8 166 30.6 33 16 24.4 197 31.2 21 17 30.4 208 27.0 29 18 25.4 349 24.0 29 19 37.8 1910 29.4 18 20 31.4 17


207

(For sample of size 5-d2 = 2.326, d3 = 0.864)

Solution: (1)

X -chart (mean chart)

Here X∑ = 587 R∑ = 420 K = 20 N = 5

CL or X :K or No.of samples

= ∑X

UCL = X= + 2

3

N

Rd = 29.35 + 3 ×

22.326

5 = 29.35 +

27.092.24

= 29.35 +12.09 = 41.44

LCL = X= – 2

3

N

Rd = 29.35 –

27.092.24

= 29.35 – 12.09 = 17.26

208


(2) R-Chart

210 205 210 212 211 209 219 204 212 209212 215 208 214 210 204 211 211 203 211

CL or R = ∑R

K =

42020

= 21

UCL = R + 3 σ R* = 21+ (3×7.802) = 21 + 23.41 = 44.41

LCL = R - 3 σ R = 21 – (3 × 7.802) = 21 – (3 ×7.802) = 21 – 23.41 = -2.41 or zero

* σ R = σ P × d3, or 2

Rd × d3=

212.326

× .864 = 9.03 × 0.864 = 7.802

Problem: 2

The following table gives the average daily production figure for 20 months each of25 working days. Giventhat the population standard deviation of daily production is 35 units, draw a control chart for the mean.

Sam

ple

Ran

ge 1

cm

= 5

O

189


209

Solution:

X -chart

Conclusion:All the points lie within the control limits. The process is well within control.

CL or =X =

4200210

20= =∑X

K

UCL = =X +

3σP

N = 210 +

3 3525

× = 210 +

1055

= 231

LCL = =X –

3σP

N = 210 –

3 3525

× = 210 –

1055

or 210 – 21 = 189

Problem 3:

15 Samples of size 4 each were taken and the observed values are given below:

Samples Observed values

1 32 20 33 6

2 42 36 52 50

3 25 15 52 63

4 22 33 34 23

5 29 30 27 31

6 30 34 26 16

7 34 31 28 34

8 11 21 20 16

9 11 22 28 31

10 36 30 35 26

11 34 16 37 26

12 27 36 51 53

13 26 35 32 37

14 25 36 37 24

15 10 28 14 13

Calculate UCL and LCL for X Chart and R chart. Also prepare the chart on graph paper. For a sample size4 the control factors are —

A2 = 0.729, d2 = 2.059, d3 = 0.880, d4 = 2.282.

210


Solution:

Samples Observed Values X or Mean R (Higest = Lowest)

1 32 20 33 06 91/4 = 22.75 33 -6 = 27

2 42 36 52 50 180/4 = 45.00 52 -36 = 16

3 25 15 52 63 155/4 = 38.75 63 - 15 = 48

4 22 33 34 23 112/4 = 28.00 34 - 22 = 12

5 29 30 27 31 126/4 = 31.50 39 - 27 = 12

6 30 34 26 16 106/4 = 26.50 34 - 16 = 18

7 34 31 28 34 127/4 = 31.75 34 - 28 =06

8 11 21 20 16 68/4 =17.00 21 - 11 = 10

9 11 22 28 31 92/4 = 23.00 31 - 11 = 20

10 36 30 35 26 127/4 = 31.75 36 - 26 = 10

11 34 16 37 26 113/4 = 28.25 37 - 16 = 21

12 27 36 51 53 167/4 = 41.75 53 - 27 = 26

13 26 35 32 37 130/4 = 32.50 37 - 26 = 11

14 25 36 37 24 122/4 = 30.50 37 - 24 = 13

15 10 28 14 13 65/4 = 16.25 28 - 10 = 18

X - Chart-

CL = 445.25

29.68Kor No.of samples 15

= =∑X

UCL = X + A2 R = 29.68 + (0.729×17.867) = 29.68+13.02 = 42.70

LCL = 16.66

FALSE BASE LINE


211

Sample number No. of defectives Sample number No. of defectives 1 6 11 10 2 2 12 4 3 4 13 6 4 1 14 11 5 20 15 22 6 6 16 8 7 10 17 0 8 19 18 3 9 4 19 23

10 21 20 10

Conclusion: Sample no. 2 is outside UCL and sample no. 15 is outside LCL.

LCL = =X – A2 R =29.68 – (0.729 x 17.867) = 16.66

R-chart –

CL or R 268

17.867Kor No.of samples 15

= =∑R =

UCL = d4×R = 2.282 × 17.867 = 40.77

LCL = d3 x R = 0 × 17.867 = 0

Conclusion : Sample 3 is outside control limits.

Problem: 4

The following table gives the result of inspection of 20 samples of 100 items each taken in 20 working days.Draw a P-chart. What conclusion do you draw from the chart about the process?

UCL = 40.77

CL = 17.86

LCL = 0

50

45

40

35

30

25

20

15

10

5

0

SAM

PLE

RA

NG

E

1 C

M =

5

2 4 6 8 10 12 14 16 18 20SAMPLE NO. 1CM = 2 SAMPLES

212


Solution:

Total No. of items inspected = No. of samples × units inspected in each sample

= 20 × 100 =2000 units

Average fraction defectives =Total no.of defectives 200

0.10Total no.of itemsinspected 2000

= =

P or CL =0.10

UCL = (1 0.10(1 0.10)

0.10 3 0.10 3 0.0009100

⎛ ⎞− −⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

P P)P +

n

= 0.10 + 3 × 0.03 = 0.10 + 0.09 = 0.19

LCL =(1 0.10(1 0.10)

0.10 3 0.10 3 0.0009100

⎛ ⎞− −⎛ ⎞= − = +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

P P)P –

n

= 0.10 – 3 × 0.03 = 0.10 – 0.09 = 0.01

Conclusion: Four points, i.e., sample numbers 5, 10, 15 and 19 lie outside the control limits.

Problem: 5

The following table gives the result of inspection of 20 samples of 100 items each taken on 20 workingdays. Draw a P-chart. What conclusion would you draw from the chart?

UCL = 19

CL = 10

LCL = 01

2422201816141210080604020

2 4 6 8 10 12 14 16 18 20

FRA

CT

ION

DEF

ECT

IVE

1 C

M =

02

SAMPLE NO. 1 CM = 2 SAMPLES


213

Conclusion: Only sample number 15 is beyond control limits.

Sample No. 1 2 3 4 5 6 7 8 9 10 No. of defectives 9 17 8 7 12 5 11 16 14 15 Sample No. 11 12 13 14 15 16 17 18 19 20 No. of defectives 10 6 7 18 16 10 5 14 7 13

Solution:

Total no. of items inspected = No. of samples × units inspected in each sample

= 20 × 100 = 2000 units.

Average fraction detectives orTotal no.of defectives 80

= 0.04Total no. items inspected 2000

= =P

Hence P or CL= 0.04

UCL = (1 ) .04(1 .04)

+ 3 0.04 3100

− −= +P PP

n = 0.04 + 3 0.000384 = 0.04 + 3×0.0196 =0.04+0.0588 =.0988

LCL = (1 ) .04(1 04)

– 3 0.04 3100

− −= −P P .P

n = 0.04 – 3 0.000384 = 0.04 – 3 × .0196 = 0.04 – 0.0588 =–0.0188=0

Problem: 6

The following table gives the result of inspection of 20 samples of 100 items each taken on working days.Draw a P-chart. What conclusion would you draw from the chart?

UCL = 09.88

CL = 04

LCL = 0

SAMPLE NO. 1 CM = 2 SAMPLES2 4 6 8 10 12 14 16 18 20

20

18

16

14

12

10

08

06

04

02

0

FRA

CT

ION

DEF

ECT

IVE

1 C

M =

02

214


Conclusion : No. point is beyond limit

UCL = 20.39

CL = 11

LCL = 0.1613

2 4 6 8 10 12 14 16 18 20

20

18

16

14

12

10

08

06

04

02

0X

Y

FRA

CT

ION

DEF

ECT

IVE

1 C

M =

02

SAMPLE NO. 1 CM = 2

Sample No. 1 2 3 4 5 6 7 8 9 10 No. of Defectives 0 2 4 6 6 4 0 2 4 8 Sample No. 11 12 13 14 15 16 17 18 19 20 No. of Defectives 8 0 4 6 14 0 2 2 6 2

Solution:

Total number of items inspected = No. of samples × units in each sample

= 20 ×100 = 2000 units

Average traction defectives or Total no.of defectives 220

0.11Total no.itemsinspected 2000

= = =P

(i) CL = 0.11

(ii) UCL = (1 ) 11(1 .11) .0979

3 0.11 3 0.11 3100 100

− − ⎛ ⎞+ = + = + ⎜ ⎟⎝ ⎠

P P .P

n= 0.11 + 3 × 0.03129 = 0.11 + 0.09387 = 0.20387 = 0.2039

(iii) LCL = (1 )

3−− P P

Pn

- 3 = 0.11 - 3 × 0.03129 = 0.11-0.09387 = 0.01613

Problem 7. 18 carpets had defects in their finish as follows. Supposing the defects follow the Poisson Law.draw a control chart for the number of defects.

No. of Defects 0 1 2 3 4 5 6

No. of carpets 0 1 2 4 3 5 3having specified No. of Defects

Solution.

C = Total no.of defectives

Total no.of carpetsexa min ed


215

= (0 0) (1 1) (2 2) (3 4) (4 3) (5 5) (6 3)

(0 1 2 4 3 5 3)× + × + × + × + × + × + ×

+ + + + + +

=0 1 4 12 12 25 18 72

or 418 18

+ + + + + + = ... CL = C = 4

UCL = C + 3 ( )C = 4 + 3 4 or 4 + 3 × 2 = 4 + 6 = 10

LCL = C – 3 ( )C = 4 – 3 ( )4 or 4 – 3 × 2 = -2 = 0

Part Nominal Dimension (cm)

Total Tolerance (cm)

A 0.600 0.002 B 0.750 0.004 C 1.250 0.006 D 0.700 0.002

Total 3.300cm 0.014cm

2 4 6 8 10 12 14 16 18 20

UCL = 10

CL = 4

LCL = 0

SAMPLE NO.

20

18

16

14

12

10

08

06

04

02

0

SCA

LE

= 1

CM

2 D

EFE

CT

SD

EFEC

TS

PER

UN

IT

Problem: 8

In the following example —

EA B C D

F0.600 + 0.750 + 0.002 1.250 + 0.003 0.700 + 0.001

Parts A, B, C and D must fit within the space EF. What should be the dimension and tolerance of EF?

Solution:

The typical engineering approach is as follows:

216


The tolerance would be half the total tolerance of 0.014 or 0.007 cm., and the dimension specificationwould be 3.3000 ± 0.007.

One statistical approach uses the formula

T t = 2 2 21 2 n.........+ + +T T T

in which T is the individual tolerance and Tt is the total tolerance. Using this concept, the tolerance forthe example would be:

Tt = 2 2 2 2(0.002) (0.004) (0.005) (0.002) 0.00775 or 0.008+ + + =

The dimension specification would then be expressed as 3.300 ± 0.004 cm. with a total tolerance of0.008 cm.

It would be noted that the statistical tolerance of 0.008 is approximately 55% of the conventionaltolerance of 0.014 cm. Now if the EF dimensions were as much as 0.014 cm or even 0.010 cm. one orseveral of the parts tolerance can be widened and still be within good practice.

Problem: 9

Incoming steel to be used in processing is tested to see that it is of the right chemical composition before itis machined. Dimensions of the machined parts are inspected, prior to the heat treating operation. Anautomatic heat treatment furnace is set at a temperature which hardens the parts. The temperature is set sothat the average force required to break the part is 32000 unit wt. The inherent variability of the heattreating process produces a standard deviation of the breaking force at 3000 unit wt. Establish the controllimits for x so that α = 0.10 when sample of size n = 4 are taken.

Solution:

30004

σσ = =n

= 1500 Unit-wt.

Z(1–2α

) = 1.645

UCL x = μ + Z σ x = 32000 + 1.645(1500) = 34467.5 unit-wt.

UCL x = μ – Z σ x = 32000 - 1.645(1500) = 29532.5 unit-wt.

Problem: 10

The radiology department of a large hospital has an average retake rate of 8.8%; i.e. 8.80% of its x-raysmust be repeated because the picture is not sufficiently clear. Errors can occur because of incorrect patientmeasurement, improper calibration or setting of the machine, poor film quality, incorrect film processingor other reasons. During the past month, 9000 x-rays were taken, and 11.2% had to be repeated. Does theprocess appear to be within its 3σ limits or does it appear that there may be some assignable cause forvariation?

Solution:

p = 0.088

xα


217

xα

4 4 5 36 2 2 45 3 4 23 2 4 55 7 5 3

Control limits for p when a sample of 9000 is taken are:

p ± 3 3(1 0.088(1 0.088)

0.088 0.09070and0.07909000 9000

− −= ± =p)

The value of 11.2% defective is very much outside the 3σ limits for the process.

Problem: 11

Twenty samples were taken from a cable-weaving machine while it is being operated under closelycontrolled conditions. The number of defects per 100 meters for the samples is recorded in the chart below.Determine the control chart limits for the machine.

Solution:

z = 78

3.920

= =∑c

n

Control limits for the number of defects per 100 meters are: c ± 3 c

UCL = 3.9 + 3 3.9 = 9.8

LCL = 3.9 – 3 3.9 = negative or, 0, whichever is greater.

Therefore LCL = 0

Problem: 12

An attribute control chart exists for part 3146B shows the average fraction defective 0.125, upper controllimit 0.200 and lower control limit 0.050, based on two months of daily data. Recently, 12 units weresampled each day for six days with units defective 2, 1, 2, 0, 3 & 3.

(a) Construct a control chart for management, carefully labelling the chart, and interpret it formanagement.

(b) What is the significance of the fraction defective for day 4 being below the lower control limit?

Solution:

The solution for part (a) is shown in the following figure. Fraction defective for the recent six days are:

2/12 = 0.166 1/12 = 0.083 2/12 = 0.1660/12 = 0 3/12 = 0.250 3/12 = 0.250

218


(b) The control limits should be set in such a way that measurements falling outside the limits areregarded as a warning signal. The lower control limits for fraction defective is useful becausesome change in methods, equipment or people has resulted in improved quality. The cause shouldbe found out.

Frac

tion

Def

ecti

ve

Sample Number

UCL

LCL

Px

x

x

x

xx0.200

0.125

0.050

MaintenanceManagement

STUDY NOTE - 3

220

Maintenance Management

3.0 Obsolescence, Replacement of Machinery

Wear and obsolescence are the two main causes for replacement of machinery in every aspect of life. Thereduction of wear is therefore a primary concern when designing appliances. Wear and tear due to passageof time and/or normal usage of plant and machinery is an accepted fact. Technological obsolescence is amajor danger which business firms face in modern era. With the development of new and better techniquesor equipment of performing a particular function, existing equipment and machines become uneconomical.Whenever a firm decides to switch over to new machines or improved product designs, existing machinedesigns are said to be obsolete. Hence, obsolescence is a major issue in the procurement and installation ofmachinery and equipment. A machine is technically obsolete when another machine can do the same jobmore efficiently, with reduced time and also at a lower cost. Technological obsolescence arises due tocontinuous improvements in the methods and techniques of production and sometime the rate ofimprovement is so fast that it becomes economical to replace the machinery before its expected life. Amachine may be replaced to reduce the running costs of the concerned machine and the new machinesproductivity will be more. In replacement decisions, the basic problem is to decide whether to replace amachine or equipment at present or at a future date. It is, therefore, necessary to determine whetherobsolescence or deterioration has reached the point where the reduction in operating costs expected fromreplacement justifies the net capital expenditure involved in installing the new machine and disposing ofthe old one.

Any function aimed at brining back or restore an item to its original or acceptable position or to keep it andretain its healthy, workable position is known as Maintenance.

Objectives of Maintenance:

The objectives of maintenance are : (i) To keep all the production facilities and other allied facilities such asbuilding and premises, power supply system, etc in an optimum working condition, (ii) To ensure specifiedaccuracy to products and time schedule of delivery to customers, (iii) To keep the down time of the machineat minimum, so that the production program is not disturbed, (iv) To keep the production cycle with in thestipulated range, (v) To modify the machine tools to meet the augmented need for production, (vi) Toimprove productivity of existing machine tools and to avoid sinking of additional capital, (vii) To keep themaintenance cost at a minimum as far as possible, there by keeping the factory Overheads at minimum,(viii) To extend the useful life of plant and machinery, without sacrificing the level of performance.

3.1 Break Down Maintenance, Preventive Maintenance, Routine Maintenance – Breakdown maintenance

Here the production facility is run without much routine maintenance until it breakdown. Once the machinebreakdown it is taken for repair and inspected to find out the defects. After identifying the defect, therequired repair is planned and the spares are procured to repair the machine. As the breakdowns arerandom in nature and the machine cannot be used during the repair period, production hours are losthence the productivity is reduced. Repair maintenance is not a recommended practice, in general, butmany a time many organizations prefer this, because they do not want to keep the machine idle formaintenance. But they ignore the fact that the break down repair costs more than the regular maintenancepractice. It is however, an economical way of maintaining certain non-critical items whose repair anddown time costs are less this way than with any other system of maintenance.


221

PREVENTIVE MAINTENANCE:

A system of scheduled, planned or preventive maintenance tries to minimize the problems of breakdownmaintenance. It locates weak parts in all equipments, provides them regular inspection and minor repairsthereby reducing the danger of unanticipated breakdowns. The underlying principle of preventivemaintenance is that prevention is better than cure. It involves periodic inspection of equipment andmachinery to uncover conditions that lead to production breakdown and harmful depreciation. The systemof preventive maintenance varies from plant to plant depending on the requirement of the plant. Anycompany, adopting the preventive maintenance should keep the record of failure of various componentsand equipment, which help the maintenance department to statistically analyze the failure pattern andreplace the item before it fails, so that the breakdown can be eliminated. This reduces the unanticipatedbreakdowns, increases the availability of the equipment, maintain optimum productive efficiency ofequipment and machinery reduces the work content of maintenance job, increases productivity and safetyof life of worker.

Production department or maintenance department depending on the size of the plant generally takes uppreventive maintenance work. As the preventive maintenance is a costly affair, it is better to maintainrecords of cost (both labour, materials used and spares used) and a valuation of the work done by thedepartment will show us what benefits are derived from preventive maintenance. The analytical approachto evaluate the work done by preventive maintenance is

(i) (Inspections incomplete) / (Inspections scheduled) × 100 should be less than 10%

(ii) (Hours worked for maintenance) / (Scheduled hours) × 100 = Performance of the department.

(iii) Down time to be given as a ratio of the available hours and to be compared against a standard to beworked out for each company or against a figure of the past. The ratio is given as:

= Down time in hours/ Available hours = working days × hours per day × number of machines. Heredown time is the total time of stoppage of the machine for scheduled and unscheduled maintenancework.

(iv) Frequency of break downs = (Number of break downs) / (Available machine hours)

(v) Effectiveness of planning = (Labour hours on scheduled maintenance) / (Total labour hours spent onmaintenance).

OR

(Down time due to scheduled maintenance)/(Down time due to total maintenance work)

Advantages of preventive maintenance are (i) Reduced breakdowns and downtime, (ii) Greater safety toworkers, (iii) Fewer large scale repairs, (iv) Less standby or reserve equipment or spares, (v) Lower unitcost of the product manufactured, (vi) Better product quality, (vii) increased equipments life and (viii)Better industrial relations.

Routine Maintenance:

It includes lubrication, cleaning, periodic overhaul; etc. This is done while the equipment is running orduring pre-planned shut-downs. Running maintenance is the work which can be carried out while thefacility is in service.

222


3.2 Maintenance Techniques

It can be discussed as under:

In some cases the loss and inconvenience due to breakdown of equipment is so high that standby equipmentis kept. As soon as the original equipment fails, the standby facility is employed to avoid interruption anddowntime. Standby machines are often kept to reduce the loss due to the breakdown of a key machine.Breakdown maintenance also requires use of standby machines. The main question here is how manystandby machines to keep and for how long. In order to decide this, a cost benefit analysis of standbymachines should be made. There are various costs involved in standby machines. First, there is interestcost on capital investment. Secondly, space is needed to keep standby machines. Thirdly, there is depreciationin the value of standby machines. Fourthly, periodic checking and servicing is necessary to keep the standbymachines in new condition. The benefits of standby machines consist of protection against a completeshutdown or shut down of operations. It avoids loss of production and, therefore, it is necessary to estimateloss of future failures a table of expected costs and benefits can be prepared.

Shifting production during breakdown. Under this method spare capacity is maintained not in the form ofstandby machines but by allowing rest to running machines at intervals and by rotation. If one machine ina production line requires shutdown, the output is maintained by shifting to under uitilised machines inother lines. For such application, the capacities of different machines must be properly matched.

3.3 Maintenance Organization

At least 50 to 60 percent of investment of any organization is spent on Building and Production facilities.Hence, it is worthwhile to give due consideration for effective maintenance of these items. The maintenancedepartment will looks after the upkeep of equipments, buildings and other. For effective contribution of itswork, the maintenance department must have proper place in the organization and it must also have agood organizational structure. While organizing a maintenance department one must remember that thereshould be clear division of authority with little or no overlap. Vertical lines of authority and responsibilitymust be kept as short as possible. Keep the span of control to an extent of 3 to 6 for a manager. Theorganizational structure should be flexible. The structure should be designed to suit the types of maintenancework involved. Depending on the need, the maintenance activity may be centralized or decentralized.

Chief Plant Engineer

Works & Service

Maintenance Workshop

Safety Engineering

Maintenance Engineering

Maintenance Stores & Spares

Management

Works & Service

Works & Service

Works & Service

Maintenance contracts

Planning

Services

Shops

Operation Preventive Maintenance Break Down

Overhauls Planning & Drawing Office

Spare Fabrication

Electrical Repair shop

Figure. Outline of a maintenance department of a large company


223

Organizing Maintenance Work

In order to facilitate proper control of maintenance work; we must enforce three rules as below.

Maintenance Request

This must be made in writing to a central point in the organization. No work should be carried out withoutthe knowledge and approval of maintenance supervision - if this discipline is not followed by theorganization, it leads to wastage of skilled manpower and inability of the maintenance personnel to scheduleessential maintenance work.

Maintenance Stores

Non-availability of vital spare parts when required to meet an emergency like breakdown, may lead toexcessive shutdown of the plant and equipment. A large number of items or materials are required to bestored and it involves investing valuable funds from the working capital. A proper stores management isessential as a backup service of good maintenance.

Records of Maintenance Work Done

Paper work for maintenance is crucial for establishing a good maintenance organization and is oftenneglected. The records of maintenance work carried out from time to time have to be kept equipment wise.History cards or logbooks of all the plants and equipment must be compiled meticulously giving details ofmaterials used, components replaced and time spent by the workforce.

Creation and maintaining this database is essential for proper planning and control, which alone will leadto effective and efficient maintenance.

To get the full benefits of effective maintenance the following requirement is to be fulfilled:

(i) Good Supervision and administration of maintenance department, (ii) Good and clear instructions to begiven to maintenance crew regarding the repair, (iii) Proper control of work in coordination with productiondepartment, (iv) Good training should be given to the maintenance personnel, (v) Good scheduledmaintenance program should be chalked out, (vi) Proper maintenance record keeping is a must, (vii) Thereshould be adequate stock of spare parts, particularly insurance spares.

3.4 Maintenance Problem

The main problem in maintenance analysis is to minimise the overall cost of maintenance without sacrificingthe objectives. There are two alternatives before management. One is to repair a machine or equipmentonly when it breaks down. This will save expense of inspection and replacement of a part before its lifetimeends. The other alternative is to replace the equipment before the expiry of its working life. This willinvolve cost of periodic shutdown for check up and repairs. However, it will avoid the loss due to suddenfailure or breakdown.

The two types of cost - cost of premature replacement and cost of breakdown - need to be balanced. Theobjective is to minimise total maintenance cost and downtime. Economic analysis is helpful in finding ajudicious combination of two types of maintenance. The relationship between preventive maintenancetime and repair time is also significant. Preventive maintenance policy is justified only when the averagedowntime and its cost is less than the average time taken to carry out breakdown repairs. If the machinehappens to be part of production line, the breakdown of a machine would throw the entire production line

224


out of gear while a preventive maintenance schedule might enable the repair to be performed during ascheduled idle time of the line.


Problem: 1

A workshop has 20 nos. of identical machines. The failure pattern of the machine is given below:-

It costs Rs. 150 to attend a failed machine and rectify the same. Compute the yearly cost of servicing thebroken down machines.

Solution:

Expected time before failure.

= 0.20×1+0.15×2+0.15×3+0.15×4+0.15×5+0.15×6 =3.5 months

Therefore number or repair/machine/annum = 12/3.5

Considering 20 machines and Rs. 150 to attend a failed machine the yearly cost of servicing

= 12/3.5×20×150 = Rs.10286.

Problem: 2

A Public transport system is experiencing the following number of breakdowns for months over the past 2years in their new fleet of vehicles:

Each break down costs the firm an average of Rs. 2,800. For a cost of Rs. 1,500 per month, preventivemaintenance can be carried out to limit the breakdowns to an average of one per month. Which policy issuitable for the firm?

Solution:

Converting the frequencies to a probability distribution and determining the expected cost/month ofbreakdowns we get:

Elapsed time after Maintenance attention

(in month) Probability of failure

1 0.20 2 0.15 3 0.15 4 0.15 5 0.15 6 0.20

Number of breakdowns 0 1 2 3 4 Number of months this occurred 2 8 10 3 1


225

Breakdown cost per month; Expected cost = 1.710×Rs.2800 = Rs.4788.

Preventive maintenance cost per month: -

Average cost of one breakdown/month = Rs.2, 800

Maintenance contract cost/month = Rs. 1,500

Total = Rs. 4,300.

Thus, preventive maintenance policy is suitable for the firm.

Problem: 3

Indian Electronics, manufactures TV sets and carries out the picture tube testing for 2000 hours. A sampleof 100 tubes was put through this quality test during which two tubes failed. If the average usage of TV bythe customer is 4 hours/day and if 10,000 TV sets were sold, then in one year how many tubes wereexpected to fail and what is the mean time between failures for these tubes?

Solution:

The total test time = (100 tubes) × 2000 hours = 200,000 tube-hours.

There are two tubes which have failed and hence the total time is to be adjusted for the number of hours

lost due to the failures during the testing.

The lost hours are computed as = 2 × 20002

= 2000 hours.

The assumption is made here is that each of the failed tubes have lasted an average of half of the testperiod.

Therefore, the test shows that there are two failures during (2,00,000 – 2000) = 1,98,000 tube hours oftesting.

During 365 days a year (four hours a day) for 10,000 tubes the number of expected failures

1,98,000

2 ×10,000 ×365 × 4 = 147.47 =148 tubes approximately.

Mean time between failures = failures 2

testingof hrs. tubes1,98,000

= 99,000 tubes hours per failure = 365 4

000,99

× = 67.8 tubes year per failure

No. of breakdowns

Frequency in months

Frequency in per cent Expected Value

0 2 0.083 0.000 1 8 0.333 0.333 2 10 0.417 0.834 3 3 0.125 0.375 4 1 0.042 0.168 Total 1.710

226


Problem: 4

A company has 50 identical machines in its facilities. The cost of preventive servicing (Cp) is Rs. 20, and thecost of repair after breakdown (CR) is Rs. 100. The company seeks the minimum cost preventive servicingfrequency and has collected the data on breakdown probabilities in the following table:

Probabilities of machine breakdown, by month:

Solution:

The mean time before failure is 5.4 months and the expected cost with no preventive maintenance

would be 100 x 4.5

50 = Rs. 925.93 per month. The following calculations show Bj, the expected number

of breakdowns between preventive maintenance intervals, for the possible intervals, that may be

considered.

B1 = MP1 =50 (0.10) = 5

B2 = m (P1+ P2) + B1P1= 50(0.10+0.05) +5(0.10) = 8

B3= 50 (0.10 + 0.05 + 0.05) + 8 (0.10) + 5 (0.05) = 11.05

Accordingly, B4 = 16.75, B5 = 25.63, B6 = 35.5, B7 = 48.72, B8 = 63.46.

Months after servicing that breakdown occurs (i)

Probability that breakdown will occur (Pi) i.Pi

1 0.10 0.10 2 0.05 0.10 3 0.05 0.15 4 0.10 0.40 5 0.15 0.75 6 0.15 0.90 7 0.20 1.40 8 0.20 1.60 1.00 5.40


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The costs of various preventive maintenance intervals are summarised in the table below :Cost of alternative preventive maintenance intervals –

A policy of performing preventive maintenance every 4 months results in the lowest average cost, aboutRs. 669. This amount is Rs. 257 per month less than the Rs. 926 expected cost without preventive maintenance.This policy would reduce the costs by (257 ÷ 926) × 100 = 27.75% below the cost of repairing the machinesonly when they breakdown.

Number of months between

preventive services (j)

Bj Expected Number of

Breakdown in j months

Expected cost/month to

Repair Breakdown

CR×Bj/j

Cost per month for preventive

service every j month CR(M)/j

Total expected cost per month of

preventive maintenance

and repair

(1) (2) (3) (4) (5) 1 5.00 500.00 1000.00 1500.00 2 8.00 400.00 500.00 900.00 3 11.05 368.33 333.33 701.66 4 16.75 418.75 250.00 668.75 5 25.63 512.60 200.00 712.60 6 35.50 591.67 166.67 758.34 7 48.72 696.00 142.86 838.86 8 63.46 793.25 125.00 918.25

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STUDY NOTE - 4

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4.0 Input-output Ratio

Input-output analysis reflects a general theory of production based on the idea of economic interdependence.Many input - output models are useful in forecasting. Input - output analysis takes into consideration theinterdependence of the different sectors in the economy. This is because; the input to one sector is outputof another sector. For example, the output of coal industry will be an input to the steel plant and the outputof steel industry is an input to the construction industry and so on. There are many such cyclic relationswithin the various sectors of economy. Taking the data of outputs and inputs and studying the relationshipbetween these two, we will be in a position to analyse the total demand for a product and the outputrequired from industrial units. This type of analysis is very important because it takes into account all theintricate relationships in the economy. One of the limitations in this method is that the utility of an outputis restricted to economic analysis, not considering the other business, governmental, technological andinternal factors. It is limited but useful analysis. The analysis need not be limited to macro-level, speakingonly in terms steel sector and coal sector etc. It may be mote ‘micro’, by considering the inputs and outputswithin a general product group in the total economy. This type of analysis is very much used and is foundmore beneficial and useful. Three major assumptions in developing this technique are:

1. The total output of an industry is consumed as input by all industries for a time period, underconsideration.

2. The input bought by each industry has usually been made dependent only on the industry’s level ofoutput.

3. The ration of an industry’s input to its output, once established is fixed. This ratio is known as input-output number or production coefficient.

Input-output analysis is a mathematical study of an economy in which different production sectors suchas agriculture, industry and services etc. have interdependence. Thus, in input-output analysis, we try toanalyse quantitatively the interdependence of inputs and outputs of various industries and find theequilibrium between the inputs and output of each industry, plant, sector or economy. The output of anyindustry depends very much on its inputs and these inputs are the outputs of other industries. This analysissignifies to foretell the total production per sector and its demand in other industries.

It can be presented as: Total output of all sectors = Total inputs of all sectors.

or, output of an industry = Total inputs of that industry.

This input-output analysis is also known as ‘analysis of inter-industry or inter-sector flows or deliveries oranalysis of inter-industry relations.’

Input-output analysis can be used in the following cases:

1. The input-output analysis can be used for the study of variation in productivity of an enterprise. Thisanalysis is based on index-numbers, using value added measure for output and the measures ofinputs all at constant prices. The value added in each industry can be obtained by subtracting thevalue of inputs used from the value of gross output.

2. Input-output table has a historic importance showing a record of past and can be a very usefulinstrument in economic analysis and framing economic policy. In developed countries, this analysis


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is used for economic forecasting and in developing countries, for economic planning or programming.In business world, business forecasting can be done by this analysis.

3. This analysis can be very usefully used for a comparative study of growth by examining. By examiningcountry-wise or sector or region-wise or plant, firm or industry-wise differences in input-outputcoefficients, the process of growth can be more easily interpreted.

4. This analysis can be utilised to determine the impact of some major exogamous changes in the economyon other factors of production.

4.1 Linear Programming

The solution technique for linear optimisation models is linear programming.

Linear programming is very important tool of operation research which has recently been started in thefield of production management. It is new mathematical technique developed while the World War II wasin progress. Before the inception of this technique in production management, every problem was tackledby trial and error method, the result of which was not so sure and best to the satisfaction. By introducingthe technique, the solution of every problem now does not depend on chance; it is now tackled with somecertainty by the use of this method.

Every business has limited resources (men materials, labour, money etc.) but wants to achieve its mainobjective of getting the maximum profit. Maximisation of profit of the business depends upon a number ofvariables so there may be a number of alternatives before the managing executives to maximise the profits.The business executives are, therefore, busy in choosing the most appropriate (best possible) action amonga number of alternative courses of action. As there are a large numbers of variables to be considered foreach alternative, the situation becomes more complicated. The management is to choose the best possiblecombination of variables. Even if information about variables is available, the problem is not so easy whenwe go in detail. It is infrequently so intricate the detail and in the interrelationships involved as to make itpossible to use in its complexity. Previously these combinations were determined by trial and error methodand no one could ascertain whether a best solution had been obtained or not. The linear programming hasmade it very simple to know about the best possible solution of any problem. Linear programming isapplied to problems which involves interaction between alternatives.

Linear programming is a recently-developed technique which is being used in production management. Itis technique of choosing a best possible alternative among the various alternatives available difficult becauseeach alternative consists of a large number of variables, giving different results. Linear programmingtechnique has considerably simplified the problem.

Linear programming consists of two words—’linear’ and ‘programming’. The word ‘linear’ establishescertain relationships among different variables whereas the word ‘programming’ indicates a way to getdesired results by the optimum use of available resources. Thus linear programming is a mathematicaltechnique for allotting limited resources in an optimum manner. Hence, it is of great practical significancefor the management for achieving the prime objective of the business, i.e., maximisation of profits. It is anundisputable fact that resources of a business are limited setting limit or restriction on the smooth functioningof the business. The problem confronting the management is to decide the manner in which these limitedresources are to be earmarked for various’ uses so as to have the maximum profit. Linear programmingtechnique solves this problem.

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Linear programming is a planning technique that permits some objective functions to be minimised ormaximised within the framework of given situational restrictions.

It is a planning technique by using the mathematical equations. It can be said as a technique of selecting thebest possible (optimal) strategy among a number of alternatives. The chosen strategy is said to be the bestbecause it involves maximisation or minimisation of some desired action, e.g., maximisation of profits orminimisation of costs.

Uses or Application of Linear Programming:

The linear programming technique is useful in the following cases:

(i) It helps in determining optimum combination of several variables with given constraints and thusselecting the best possible strategy among various alternatives available.

(ii) Linear programming provides additional information for proper planning and control over variousoperations in the organisation.

(iii) The management must understand the activities of the organisation for constructing suitablemathematical model visualising the relationship between variables, if any, and making improvementover them. The linear programming helps in better understanding the phenomenon.

(iv) Linear programming contributes to the development of executives through the techniques of modelbuilding and their interpretations.

(v) Linear programming provides standards for the management problems by defining (a) the objectivesto be pursued, (b) various restrictions to be imposed, (c) various alternatives available and relationshipbetween them, (d) the contribution of each alternative to the objectives.

Application: The linear programming technique may be fruitfully applied in the following spheres:(i) The linear programming can be used in production scheduling and inventory control so as to produce

the maximum out of the resources available to satisfy the needs of the public by minimising the costof production and the cost of inventory control.

(ii) The technique of linear programming can also be fruitfully used in solving the blending problems.Where basic components are combined to produce a product that has certain set of specifications andone may calculate the best possible combination of these compounds which maximise the profits orminimise the costs.

(iii) Other important applications of linear programming can be in purchasing, routing, assignment andother problems having selection problems such as(a) selecting the location of plant, (b) deciding thetransportation route within the organisation, (c) utilising the godowns and other distribution centresto the maximising, (d) preparing low-cost production schedules, (e) determining the most profitableproduct-mix , and (f) analysing the effects of changes in purchase prices and sale prices.

Limitations of Linear Programming: Linear programming, as the name suggests, presents the relationsbetween different variables in a line. This presumption, however, is not correct. There are many limitationsof linear programming which can be given below:

(i) The technique solves the problems of linear nature whereas, in practice, most of the business problemsare of non-linear nature. The solution of non-linear problems is not possible through this technique.


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(ii) This technique cannot provide solutions to problems which involve variables not capable of beingexpressed quantitatively.

(iii) Under this technique, uncertainties are not considered whereas in business, every problem is full ofuncertainties.

(iv) One of the limitations of linear programming is that the results under this technique are not necessarilyto be in whole numbers. Sometimes results in points give a misleading picture. For example, if itresults by this technique that 1.6 machines are to be purchased, it becomes very difficult to ascertainwhether to purchase one machine or two machines because machines cannot be purchased in fraction.

Formulation of an L.P problem: In formulating the linear programming problem, the basic step is to setup some mathematical model. For this purpose, the following considerations should be kept in mind: (i)unknown variables, (ii) objectives and (iii) constraints. This can be done with the help of the followingterms:1. The objective function: An objective function is some sort of mathematical relationship between

variables under consideration. Under linear programming this relationship is always linear. Theconstruction of objective function mainly depends upon abstraction. It is a process whereby the mostimportant features of a system are considered.

The objective function is always positive. The coefficients a1, a2 are certain constants and are known asprices associated to the variables X1, X2.

2. Constraints on the variables of the objective function: In practice, the objective function is to beeoptimised under certain restraints imposed on die variables or some combination of few or all thevariables occurring in the objective function. These restrictions, in most of the cases, are never exact.Had these been exact, the objective function would have been easily optimised by die use of differentialcalculus. The restraints should be known and expressed in terms of linear algebraic expression.

3. Feasible solution: One of the essential features of linear programming problem is optimisation oflinear objective function Z. It is subject to the linear constraints on the variables of the objective function.A set of values of X1, X2………which satisfies the constraints and the non- negativity restrictions arecalled Feasible solution. A feasible solution which optimises the objective function is known asOptimalsolution. Thus, a linear programming problem can be formulated in this way.

Simplex Method of Linear Programming

There are number of ways of finding the optimal solution for a given linear programming problems. Thefollowing three methods are mainly used for this purpose.

(1) Graphic Method

(2) Simplex Method

(3) Transportation Method.

1. Graphic method. This method is generally used for solving the problems having two or three variables.Due to this limitation of handling only two or three variables at a time this method has limitedapplication in industrial problems. In practice, two variable cases are easy to solve by this methodbecause three dimensional geometry becomes too complicated to find accurate results.

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2. Simplex Method. This is the most powerful and popular method for solving linear programmingproblems. Any problem can be solved by this method which satisfies the conditions of linearity andcertainty irrespective of the number of variables. It is an iterative procedure which ultimately givesthe optimal solution.

3. Transportation Method. This method is used to know the minimum cost of transportation of a productfrom various origins to different distribution and consumption centres.

Simplex method of linear programming. Simplex method of linear programming is a very importanttechnique to solve the various linear problems. Under this method, algebraic procedure is used to solveany problem which satisfies the test of linearity and certainty. It is an iterative procedure which ultimatelygives the optimal solution. Several variables can be used under this method. However, the simplex methodis more complex and involves somewhat unsophisticated complex mathematics.

Features. There are two characteristic features of the simplex method which we can cite them at the veryoutset.

First, we have the computational routine in the iterative process. Iteration means the repetition. Therefore,in working towards the optimal solution, the computational routine if repeated again and again, followinga standard pattern, successive solutions are developed in a systematic pattern until the best solution isarrived.

Secondly, such new solution will give a profit larger than the previous solution. This important characteristicassures us that we are always moving closer to the optimal solution.

There are two types of linear programming problems to be solved by this method—

(1) Maximisation of objective function, and (2) Minimisation of objective function.

1. Maximisation of objective function. The following procedural steps are taken for getting the optimumsolution by simplex method, which maximise the objective function. We shall explain the procedure ofmaximisation of objective function by simplex method with the help of the following illustration.

Illustration. A small manufacturing firm produces two types of gadgets, A and B, which are first processedin the foundry, then sent to the machine for finishing. The number of man-hours of labour required in eachshop for the production of each unit of A and B and the number of man-hours the firm has available perweek are as follows:

Foundry Machine shop Gadget A 10 5 Gadget B 6 4 Firm’s capacity per week (in hours) 1000 600

Construct the objective function and the corresponding constraints for calculating that how many unitsshould be produced per week so that the profit is maximum. The profit on the sale of A is Rs. 30 per unit ascompared to B’s Rs. 20 per unit.

Taking this illustration, we can explain the following steps:

1. The first step is to know the objective function and the linear constraints for the given problem.

2. Slack variables. The next step is to change the inequalities for two constraints into equations. Theinequalities are converted into equations only by the device of introducing slack variables. A slack


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variable represents costless process whose function is to ‘use up’ otherwise unused capacity, saymachine time or warehouse capacity. Effectually, the slack variable represents unused capacity, andit will be zero, only if production facilities or capacities are fully utilised. In each constrained equation,the variables used in other equations are also introduced but with zero coefficients. The same slackvariables are also introduced in the objectives function but by the time it is maximum, all their coefficientwill be zero. The equation can be developed as follows:

Z = 30X + 20Y + 0S1 +0S2;

Subject to:

10X + 6Y + 1S1+ 0S2 =1000

5X + 4Y +0S1 +1S2 = 600

Here, S1 and S2 are slack variables (unused time on machine 1 and 2 respectively). Slack variables arealways non-negative.

3. Developing initial simplex tableau.We can now set out this whole problem in a simplex tableau andread it for a basic feasible solution. This table consists of rows and columns of figures and is alsoknown a simplex matrix. Each table has the following columnwise heading from left to right:

(i) (ii) (iii) (iv) (v) (vi) Unit profit

(Cj) Basis or

programme Quantity or

Constant Slack

Variables Real Variables Ratio

The following entries are made in each column as:

(i) The first column (unit-profit column) has zero entry in each row in the initial stage.

(ii) The second column (basis or programme column) shows the first feasible solution and containsvarious slack or artificial variables from top to bottom. The first feasible solution consists ofthose slack or artificial variables for which the corresponding column is an identity vector.

(iii) The third column (the quantity or constant column) will have the constant right hand side valueof each constrained equation in the respective row.

(iv) The fourth column (slack variables column shall have the respective coefficients of the givenslack variables in different constrained equations for each row. These coefficients will be either 1or 0 depending whether the slack variables of the column appears in the corresponding constraintequation or not.

(v) The fifth column (real variables column) shall have the respective coefficients of the given realvariables in different constraint equations for each row.

(vi) The six and last column (ratio column).

Now with the help of the given illustration, we can prepare the initial simplex tableau, illustrating theabove steps:

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Initial Simplex Tableau

j 1 2 3 4 5 6 Cj 0 0 0 30 20 Constant S1 S2 X Y Ratio

Unit profit

Progr- amme

0 S1 1000 1 0 10 6 100 Key Row 0 S2 600 0 1 5 4 120 Base Zj 0 0 0 0 0 Row Values Zj-Cj 0 0 0 -30 -20 Key

Column

Negative largest values

Top j 1 2 3 4 5 6 Cj 0 0 0 30 20

Row No.(l) Unit Profit

Programme Constant Slack Variables Real Variables Ratio

S1 S2 X Y 1 0 S1 1000 1 0 10 6 2 0 S2 600 0 1 5 4

It can be observed from the above table that the entries corresponding to the columns for slack variablesS1, and S2 are (1,0) and (0,1) respectively. These are unit vector. Hence the programme column of theabove table contains the slack variables S1 and S2 as its first feasible solution.

Thus Row 1 corresponds to constraint equation 10X + 6Y + S1 + 0.S2 =1000

and Row 2 corresponds to the constraint equation 5X + 4Y + 0S1 + 1S2 =600

4. Computation of base row. Having prepared the initial simplex tableau, the next step in the preparationof base row values from the initial tableau by the following procedure to get tableau I:

(i) Multiply each row entry by the corresponding entry of unit profit column.

(ii) Find the sum of the products obtained in step for each column and denote it by Zj’s.

(iii) Subtract the corresponding objective factor Cj of top row from Zj. The base row values can bepositive, zero or negative.

The Zj’s and the base row values for the above illustration from the initial simplex tableau are calculatedas below:

Z1 = (1000×0) + (600×0) = 0 and (Z1- C1) = 0-0 = 0

Z2 = (1×0) + (0×0) = 0 and (Z2 - C2) = 0 - 0 = 0

Z3 = (0 × 0) + (l× 0) = 0 and (Z3 – C3) = 0 - 0 = 0

Z4 = (10 ×0) + (5 × 0) = 0 and (Z4–C4) = 0 -30 = -30

Z5 = (6 × 0) + (4 × 0) = 0 and (Z5–C5) = 0 - 20 = -20

The simplex tableau now will be

Tableau I


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5. Examining base row values. Examine all base row values in Tableau I. If they (all values) are positive,it is the optimum solution and the solution can be read from constant and programme columns. If allvalues (Zj-Cj) are negative, the largest numerically negative value among the base row values shouldbe searched. The column in which the largest value lies is known as key column. In the above case,column 4 is the key column. And now we should proceed to step (6) and (7) below.

6. Computation of ratio column. Having located the key column, the next step is to compute the ratiocolumn. This may be done by dividing the constant column value of each row by key column entry ofthat row. So, in Tableau I, the ratio column comes to 1000/10 and 600/5.

Now examine the ratio-column of tableau to locate the key row.

The row for which entry in ratio column is non-negative and numerically smallest is taken as the keyrow. For example, in Tableau I the least non-negative entry in the ratio column is 100 hence Row I isthe key row. Negative values in the ratio column are not considered because solution values of realvariables cannot be negative.

The entry lying in the cell at the intersection of key row and key column is known as key number, e.g.,in Tableau I, 10 is the key number.

7. Preparation of Tableau II and Tableau III. After identifying key row, key column and key number inTableau I in step (6), Tableau II is prepared having some columns but with the following changes :

(i) In Tableau II, the slack variables lying in the programme column of key row of tableau I isreplaced by variable of the key column in tableau I, e.g., in tableau II of the above example, theentry in programme column of key row is obtained by replacing the slack variable S1 in theprogramme column of the key row in tableau I by the variable Y.

(ii) The value of Cj lying in the key column of Tableau I is entered in profit column of key row intableau II, e.g., 30 is written in profit column of Row I.

(iii) The remaining entries of the row in tableau II corresponding to key row in tableau I are obtainedby dividing the corresponding entries of key row in tableau I by key number, e.g., the entry infirst row and constant column of tableau II shall be 100 (100/10). Similarly, other entries in Row;I are (1/10, 0/10, 10/10 and 6/10). Now the tableau II will be as under:

Tableau II

j 1 2 3 4 5 6 Cj 0 0 0 30 20 Unit

profit Progr-amme

Constant S1 S2 X Y Ratio

1 30 X 100 0.5 0 1 0.6 167 key number

2 0 S2 100 -0.5 1 0 1 100 Key row

Zj 3000 3 0 30 18 Zj-Cj 3000 3 0 0 -2 Key column

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(iv) The entries in Row 2 of tableau II are made in the following way : The entry for any column inRow II of Tableau II is obtained by multiplying the entry in the row of Tableau II obtained in step(iii) above by the value lying in key column of row in tableau I and then subtracting the productfrom the corresponding entry of the column in Row II of Tableau I, e.g., 100 in constant columnof Row II in tableau II is equal to 600 - (100 × 5) = 100.

Similarly other entries in Row II in Tableau II are

[(0 - (0.1 × 5) = - (0.5)], [(1- (0 × 5) - 1)], [(5 - (1 × 5) =0)], [(4 - (0.6 ×5) =1)].

The base row in Tableau II is obtained by applying the operations in step (4) on the values ofTableau II and it is examined whether the optimal solution is obtained or not. If optimal solutionis reached, we shall stop at this point. If not, operations in step (6) and (7) will be repeated on theentries in Tableau II to get the tableau III and so on the procedure is repeated till we get theoptimal solution.

It is observed from Tableau II of the given problem that only Zj-Cj is negative. Hence the columnNo. 5 of tableau II is the, key column. Similarly from the ratio column of Tableau II we find theentry 100 in row II is minimum hence Row II become the key row and I is the key number.

The Tableau III will now be prepared as

Tableau III

In Tableau III all Zj-Cj are non-negative. Hence the optimal solution is reached. From programme andconstant columns of Tableau III, the result is read as X = 40, Y = 100 with minimum profit of Rs. 3,200.

2. Minimisation of objective function. The second problem of simplex method is minimisation of objectivefunction. For this purpose, the problem is first changed to maximisation problems by writing Z* = -Z = -(aX + bY)

Thus, the constraint remains unchanged. The maximum value of Z* is determined on the same lines asdescribed in maximisation problem. Maximisation of Z* implies minimisation of Z.

Degeneracy Method. Sometimes, it happens that during the course of simplex procedure, we get two ormore entries in the ratio column of any tableau to be identical and minimum. Now, the question ariseswhich row should be taken as key row? The selection of key row determines the variable to be deleted.This is known as degeneracy. The problem of degeneracy may also occur when one of the constraints onthe right hand side of the equation is zero.

The degeneracy is resolved by the following procedure:

(a) Each element of the row which have identical entries in the ratio column is divided by the key columnnumber of the respective row.

(b) The values so obtained are compared step by step from left to right. Priority is to be given to identify

Unit profit

Cj Programme

0 Constant

0 S1

0 S2

30 X

20 Y

30 X 40 0.4 -0.6 1 0 20 Y 100 -0.5 1 0 0 Zj 3,200 2 2 30 20 Zj-Cj 3,200 42 2 0 0


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columns corresponding to slack and artificial variables. The remaining columns are considered fromleft to right.

(c) The comparison is stopped as soon as the rows yield unequal ratios. The row having algebraicallysmaller ratio is taken to be the key row.

After selecting the key row, regular simplex procedure is resumed.

Slack and Artificial Variables

Slack variables.

A slack variable represents costless process whose function is to ‘use up otherwise unemployed capacity,say, machine hour or warehouse capacity. Effectually the slack variable represents unused capacity and itwill be zero only, if production facilities are fully utilisesd. There are always non-negative or positive andexplains the unallocated portion the given limited resources. The main purpose of introducing slack variablesin the simplex method of linear programming is to convert the inequalities(i.e., constraints); into equalitiesor, say, equations.

Artificial variables: The artificial variables are fictitious and do not have any physical meaning. Thesevariables are assigned a very large per unit penalty in the objective function. The penalty is designated as-- M for maximisation problem and +M in minimisation problem where M is always greater than zero.

There may be situations where the constraints involve mixture of > = and < signs. In such cases, the slackvariables cannot provide a starting feasible solution and we have to introduce both slack and artificialvariables to have a starting basis. The artificial variables are denoted as A1,A2………….An in the programmecolumn of initial simplex tableau. These artificial variables should be removed first by using simplexcriterion. If at any stage of simplex procedure the basic feasible solution contains artificial variable but thevariable to be deleted at that stage is not the artificial variable, it means we are departing from the simplexcriterion for selecting the variable.

The iterative procedure does not work under following circumstances -

(i) One or more artificial variables appeals in the program column at a stage when the base row valuesare all positive.

(ii) No artificial variables remain in the programme column.

(iii) The problem is redundant if one or more artificial variables occur in programme column and thecorresponding values in the constant column are zero at a stage when all Zj-Cj are positive or non-negative.


Problem 1. Find the non-negative values of X1, X2 and X3 that maximise the expression

Z=3X1+5X2+4X2

subject to the following restraints

2X1 + 3X2< 8

2X2 + 5X3<10

3X1+ 2X2+4X3<15

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Solution :

By introducing the slack variables to change the restrictions into equalities, we get;

2X1+ 3X2 + S1 = 8

2X2 + 5X3 + S2 =10

3X1+ 2X2+4X3 + S3= 1

Now, objective function becomes

Z = 3X1 + 5X2 +4X3 + 0S1 + 0S2 + 0S3

Now various tables are prepared to get the non-negative values of X1, X2 and X3 to maximise Z in thefollowing way:

Tableau I

Cj 3 5 4 0 0 0 Profit Programme Constant X1 X2 X3 S1 S2 S3 Ratio

0 S1 8 2 3 0 1 0 0 2.67 0 S2 10 0 2 5 0 1 0 5 0 S3 15 3 2 4 0 0 1 7.5 Zj 0 0 0 0 0 0 0

Cj-Zj - 3 5 4 0 0 0

Tableau II


5 X2 8/3 2/3 1 0 1/3 0 0 0 S2 14/3 - 4/3 0 5 -2/3 1 0 0 S3 20/3 5/3 0 4 -2/3 1 1 14/15 Zj 40/3 10/3 5 0 5/3 0 0 29/12

Cj-Zj - - 1/3 0 4 -5/3 0 0

Tableau III


5 X2 8/3 2/3 1 0 1/3 0 0 4 4 X3 14/15 -4/15 0 1 -2/15 1/5 0 -7/2 0 S3 89/15 41/15 0 0 -2/15 -4/5 0 89/41 Zj 34/15 5 4 17/15 4/5 0 0

Ci-Zj - 11/15 0 0 -17/15 -4/5 0

Tableau IV


5 X2 150/123 0 1 0 45/123 24/123 0 4 X3 930/615 0 0 1 -90/615 75/615 0 3 X1 89/41 1 0 0 -2/41 -12/14 0 Zj 765/41 3 5 4 45/41 24/41 0

Ci-Zj - 0 0 0 -45/41 -24/41 0

≥


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Thus, in Table IV all Cj-Zj are zero or negative hence it is the optimal solution.

In this, X1=89/41; X2 =150/123 and X3 =930 /615 or 62/41

and maximum profit Z = 765/41

Problem 2. Find the maximum value of Z = 3X1+ 5X2 + 4X3. Where X1, X2, X3 ≥ 0 subject lo the followingconstraints.

2X1 + 3X2 < 18

2X2+ 5X3 <18

3X1 + 2X2 + 4X3 < 25

Solution:

As here the objective function is to be maximised, we shall introduce the slack variables in theconstraints and the objective function and we get the following equations:

2X1 + 3X2+S1 + 0.S2 + 0.S3 =18

2X2+5X3 + 0.SI + S2 + 0.S3=18

3X1 + 2X2 +4X3 + 0.S1 +0.S2 +S3 = 25

So, Z = 3X1+ 5X2 + 4X3 +0S1+ 0S2 + 0S2

Now, various simplex tableaus are prepared to get the optimal solution.

Tableau I

Cj 0 0 0 3 5 4 Profit Programme Constant S1 S2 S3 X1 X2 X3

Ratio

0 S1 18 1 0 0 2 3 0 6 0 S2 18 0 1 0 0 2 5 9 0 S3 25 0 0 1 3 2 4 12.5

Zj 0 0 0 0 0 0 0 Zj – Cj — 0 0 0 –3 –5 –4

Tableau II


Ratio

5 X2 6 1/3 0 0 2/3 1 0 0 S2 6 -2/3 1 0 4/3 0 5 6/5 0 S3 13 -2/3 0 1 5/3 0 4 13/4 Zj 30 5/3 0 0 10/3 0 0 Zj – Cj - 5/3 0 0 1/3 0 -4

Tableau III


Ratio

5 X2 6 1/3 0 0 2/3 1 0 9 4 X3 6/5 -2/15 1/5 0 -4/15 0 1 Neg. 0 S3 41/5 -2/15 -4/5 1 41/15 0 0 3

Zj 74/5 17/15 4/5 0 34/15 5 4 Zj – Cj – 17/15 4/5 0 11/15 0 0

≥

α

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In Tableau IV all Zj - Cj are non-negative or positive hence it is the optimal solution. The optimal solutionis X1 = 3; X2 = 4; X3 = 2 and the maximum profit Zj = 37.

Problem 3. Maximise 3X1 + 2X2 under the following restrictions:

X1 > 0, X2 > 0

2X1+X2 < 40

X1+X2 < 24

2X1+3X2 < 60

Solution. By introducing slack variables in constraints, we get

2X1 + X2 + S1 +0S2 + 0S3 = 40

X1 + X2 + 0S1+ S2 + 0S3 = 24

2X1 + 3X2 + 0S1 + 0S2 + S3 = 60

and the objective function (Z) now can be represented by

Z = 3X1 + 2X2 + 0S1+ 0S2 +0S3

Now, the following tableaus are to be prepared to get the optimal solution:

Tableau IV

Programme Cj 0 0 0 3 5 4

Profit Constant S1 S2 S3 X1 X2 X3 Ratio5 X2 4 15/41 8/41 -10/41 0 1 04 X3 2 -6/41 5/41 4/41 0 0 13 X4 3 -2/41 -12/41 15/41 1 0 0

Zj 37 45/41 24/41 15/41 3 5 4Zj - Cj 45/41 24/41 15/41 0 0 0

Tableau I

C1 3 2 0 0 0

Profit Programme Constant X1 X2 S1 S2 S3 Ratio

0 S1 40 2 1 1 0 0 20

0 S2 24 1 1 0 2 0 24

0 S3 60 2 3 0 0 1 30

Zj 0 0 0 0 0 0

Cj – Zj 3 2 0 0 0


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Since all Cj-Zj are either zero or negative, it is the optimal solution of the problem. Here X1,

= 16; X2 = 8 and maximum profit Zj = 64.

Problem 4. A company manufactures two items X1 and X2. They are sold at a profit of Rs. 30 per unit of X1

and Rs. 20 per unit of X2. X1 requires 2kgs of materials, 3 man-hours and 1 machine hour per unit. X2

requires 1 kg of material, 2 man hours and 3 machine hours per unit.

During each production run there are 280 kgs of material available, 500 labour hours and 420 hours ofmachines used. How much of the two items should the company produce to maximize profits?

Maximise 30X1 + 20X2 subject to

2x1+x2 < 280

3xl + 2x2 < 500

x1 + 3x2 < 420

x1, x2, x3 > 0

Solution.

The objective function to be maximised in this problem may be expressed in the following way:

Z = 30X1+ 20X2

Subject to constraints

2x1 + x2 < 280

C1 3 2 0 0 0


3 X1 20 1 1/2 1/2 0 0 40

0 S2 4 0 1/2 –1/2 1 0 8

0 S3 20 0 2 –1 0 1 10

Zj 60 3 3/2 3/2 0 0

Cj – Zj — 0 1/2 –3/2 0 0

Tableau II

C1 3 2 0 0 0


3 X1 16 1 0 1 –1 0

2 X2 8 0 1 –1 2 0

0 S3 4 0 0 1 –4 1

Zj 64 3 2 1 1 0

Cj – Zj — 0 0 –1 –1 0

Tableau III

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3x1 + 2x2 < 500

x1 + 3x2 < 420

x1, x2, x3 > 0

Now, by introducing slack variables, we get

2X1 + X2 +S1 + 0S2 + 0S3 =280

3X1 + 2X2 + 0S1 + S2 + 0S2 = 500

X1 + 3X2 + 0S1 + 0S2 +S3 = 420

and Z = 30X1 + 20X2 + 0S1+ 0S2 +S3

Now, various tables will be prepared as under:

Tab leau I

Cj 30 20 0 0 0 Profit Programme Constant X1 X2 S1 S2 S3 Ratio

0 S1 280 2 1 1 0 0 140 0 S2 500 3 2 0 1 0 166.67 0 S3 420 1 3 0 0 1 420

Zj 0 0 0 0 0 0 Cj-Zj 30 20 0 0 0

Tab leau II


30 X1 140 1 1/2 1/2 0 0 280 0 S2 80 0 1/2 -3/2 1 0 160 0 S3 280 0 5/2 -1/2 0 1 112

Zj 4200 30 15 15 0 0 Cj-Zj 0 0 5 -15 0 0

Tab leau III


30 X1 84 1 0 3/5 0 -1/5 0 S2 24 0 0 -7/5 1 -1/5

20 X2 112 0 1 -1/5 0 2/5 Zj 4760 30 20 14 0 25/2 Cj-Zj — 0 0 -14 0 -25/2

In the above Tableau III, all Cj-Zj are zero or negative, hence it is optimal solution X1 = 84 ; S2

= 112 and maximum profit (Zj) = 4,760.

Problem 5. An animal feed company must produce 200 lbs. of a mixture containingthe ingredients X1 and X2. X1costs Rs. 3 per Ib. and X2 costs Rs. 8 per Ib. Not more than 80 lbs of X1can beused and minimum quantity to be used for X2 is 60 lbs. Find how much of each ingredient should be usedif the company wants to minimise the cost.


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Solution:

In the above problem, the objective function to be minimised is—

Z = 3X1 + 8X2, subject to the following constraints—

X1 < 80; X1 + X2 = 200; X2 > 60.

As, in this problem, the constraints contain all the symbols, < , = , > , so slack and artificial variables tochange the restrictions into equalities and we get—

X1 + S1 = 80;

X1 + X2 +A1 = 200 and

X2 – S2 + A2 = 60

Hence the objective function becomes—

Z* = -Z = -3X1 -8X2 + 0S1 + 0S2 - M.A1 -M.A2

Here Z * is to be minimised, hence penalty for A1 and A2 in the objective function will be -M. Thevarious tableaus under simplex method will be prepared in the following manner -

Tableau I

Cj 0 0 -M -M -3 -8 Profit Programme Constant S1 S2 A1 A2 X1 X2 Ratio

0 S1 80 1 0 0 0 1 0 -M A1 200 0 0 1 0 1 1 200 -M A2 60 0 -1 0 1 0 1 60 Zj 0 M -M -M -M -2M

Zj - Cj 0 M 0 0 3-M 8-2M

Tableau II


0 S1 80 1 0 0 0 1 0 80 -M A1 140 0 0 1 -1 1 0 100 0 X2 60 0 1 0 -1 0 1 Zj 0 -M+8 -M M-8 -M -8

Zj - Cj 0 -M+8 0 2M-8 -M+8 0

Tableau III


-3 X1 80 1 0 0 0 1 0-M A1 60 -1 1 1 -1 0 0 60 -8 X2 60 0 -1 0 1 0 1 -60 Zj -1200 5 0 0 0 -3 -8

Zj - Cj 5 0 M M 0 0

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Here all ZJ-C

J in Tableau IV are positive, hence it is the optimal solution. This is given X1= 80,

X2 = 120, the minimum value being Rs. 1200.

Problem: 6

The annual hand made furniture show and sales occurs next month and the school of vocational studies isplaying to make furniture for the sale. There are three wood working classes - I year, II year, III year at theschool and they have decided to make three styles of chairs A, B and C. Each chair must receive work ineach class and the time in hours for each chair in each class is given.

Chair I year II year III year A 2 4 3 B 3 3 2 C 2 1 4

In the next month there will be 120 available in I year class, 160 in the second year class and 100 hours inthird year class to produce chairs. The teacher of the wood working class feels that a maximum of 40 chairscan be sold at the show. The teacher has determined that the profit from each type of chair will be A — Rs.40, B — Rs. 35 and C — Rs. 30.

Formulate a linear programming model to determine how many chairs should be produced to maximiseprofit.

Solution:

Let x1, are the chairs produced of A type

x2 are the chairs produced of B type

x3 are the chairs produced of C type

Formulation

2x1 + 3x2 +2x3< 120

4 x1 + 3x2+ x3 < 160

3x1 + 2x2 + 4x3< 100

x1, x2, x3 > 0

Objective function

Maximise Z = 40 x1+35 x2+30 x3

The problem is solved by simple method, convert inequalities into equalities by adding slack variables

S.T.

2x1 + 3x2 + 2x3+ xa = 120

4x1 + 3x2 + x3 + xb= 160

3x1 + 2x2 + 4x3 + xc = 100

Rewriting objective function as Z – 40 x1 - 35x2 - 30x3 = 0Selection of non-basic variables n – m = 6 - 3 = 3where n = No. of variables

m = No. of constraint equation, where xa, xb and xc are slack variables.

∴


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Non basic variables xl = x2 = x3 = 0 gives.

∴ xa = 120

∴ xb = 160

∴ xc= 100

The initial solution is represented in starting table.

Starting table

∴

Basics Z x1 x2 x3 xa xb xc RHS Ratio Z 1 -40 -35 -30 0 0 0 0 - xa 0 2 3 2 1 0 0 120 60 xb 0 4 3 1 0 1 0 160 40 xc 0 3 2 4 0 0 1 100 33.33

Entering variable x1 leaving variable = xc literation no. 1

Basics Z x1 x2 x3 xa xb xc RHS Ratio Z 1 0 - 25/3 70/3 0 0 40/3 4000/3 - xa 0 0 5/3 - 2/3 1 0 - 2/3 160/3 32 xb 0 0 1/3 - 13/3 0 1 - 4/3 80/3 80 xc 0 1 2/3 4/3 0 0 1/3 100/3 50

Entering variable x2 leaving variable = xa literation no.2

Basics Z x1 x2 x3 xa xb xc RHS Ratio Z 1 0 0 20 5 0 10 1600 - xa 0 0 1 - 2/5 3/5 0 - 2/5 32 xb 0 0 0 42/10 - 3/25 1 - 94/75 20 xc 0 1 0 19/9 - 2/5 0 17/15 12

As Z row has all positive values hence optimal solution has reached

∴ Z = 1600

x1= 12 units

x2= 32 units

x3= 0

Problem 7.

A company has three plants F1, F2 F3 from which it supplies to 4 markets: ABCE. Determine the optimaltransportation plan from the following table giving the plant to market shipping costs, quantities available

Plant

Markets

A B C D Available at plant F1 13 11 15 20 2 F2 17 14 12 13 6 F3 18 18 15 12 7

Requirement 3 3 4 5 15

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at each plant and quantities required at each market.

Solution:

The problem is balanced transportation problem because the demand and supply both are equal to15. Now obtain the initial feasible solution by Vogel’s method. The cost matrix and the penalties for

given problem are as under:

Now, penalty of first column is maximum and the lowest cost in column (1,1), i.e., Rs. 13, so allocate 2units (maximum quantities available in F,) to cell (1,1) and delete row (1) as its requirements is complete.

Now penalty of column (2) is maximum and the lowest cost is 14 so, we can allocate 3 units to cell (2,2)as 3 units are the maximum requirement for market B. The requirement of column (2) is exhaustedhence column (2) is deleted. The shrunken matrix becomes as—

W arehouses M arkets Row Factory

A B C D Penalty

(1) (2) (3) (4)

A vail-

A bility

F1 13 11 16 20 2 (13 –11) =

2

F2 17 14 12 13 6 (13 – 12) =

1

F3 18 18 15 12 7 (15 – 12) =

3

A m ount

required

3 3 4 5 15

C olum n

penalty

(17–13)

= 4

(14–11)

=3

(15 - 12)

=3

(13 – 12)

=1

— —

Factory Warehouse Availability Penalty (1) (3) (4) F2 17 12 13 3 1 F3 18 15 12 7 3 Requirement 1 4 5 10 Penalty 1 3 1

Now, we consider earlier row(3)or column(3)because both have minimum cost of Rs. 12. Let us considercolumn (3) and allocate 3 units to cell (2,3) and delete row F2. Now the balance will be allocated to rowF3. Thus, the initial solution by Vogel’s method is given in the following table:

Factory Warehouse Availability A B C D F1 2 – – – 2 F2 – 3 3 – 6 F3 1 – 1 5 15 Requirement 3 3 4 5 7

Here m+n-1 =6 equal to the occupied cells hence solution is feasible. Now, the optimality can betested by calculating the implicit cost for each cell. The calculations are given below:


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In the above table in cell 1,2, the implicit cost is greater than the actual cost hence 8 units’ are allocated to(we get the following solution):

Factory Warehouse Availability A B C D

13 11 15 20 F1 13 11 9 7 2 –5

17 14 12 13 2 4 6 –2 F2 16 14 12 10

18 18 15 12 F3 2 29 5 6 12 15 14 12 Requirement 3 3 4 5 15 Vj 18 16 14 12

In the above table, the minimum of negative θ is one hence take θ =1 and get the revised solution and testthe optimality again with the help of the following table:

Factory Warehouse Availability A B C D F1 2– θ θ – – 2 F2 – 3– θ 3+ θ – 6 F3 1– θ – 1– θ 5 7 Required 3 3 4 5 15

Warehouse Factory A B C D

Availability

13 11 15 20 2 –5 F1 13 12 10 17

2

17 14 13 3 –3 F2 15 14 9

6

16 18 15 12 5 0 F3 18 17 15 12

7

Vj 18 17 15 12 Required 3 3 4 5 15

Thus, it can be observed from the above table that implicit cost for all cells is less than the correspondingactual costs. Hence the improve solution is optimal and the minimum cost is Rs. 156 as under

Transportation cost

F1A = 11 × l = 13

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F1B = ll × 1 = 11

F2B = 14 × 2 = 28

F2C = 12 × 4 = 48

F3A = 18 × 2 = 36

F3D = 12 × 5 = 60

Rs. 196

Problem 8. The Raja Company has two factories A and B located at some distance and three regionalwarehouse R, S, T. The transportation manager must schedule shipments for the coming week accordingto the following:

Warehouse R requires 70 tonnes

Warehouse S requires 60 tonnes

Warehouse T requires 50 tonnes

Capacity of factory A—100 tonnes

Capacity of factory B—200 tonnes

Transportation costs are as follows:

From factory A to warehouse R — Rs. 30 per tonne

From factory A to warehouse S — Rs. 10 per tonne

From factory A to warehouse T — Rs.50 per tonne

From factory B to warehouse R — Rs. 20 per tonne

From factory B to warehouse S — Rs. 40 per tonne

From factory B to warehouse T — Rs. 60 per tonne

Find the least cost shipping schedule.

Solution:

In the given problem, the total quantities available are 300 tonnes (100 tonnes in factory A and 200 tonnesin factory B) whereas the total requirements for all the three warehouses are only 180 tonnes (70 tonnes +60 tonnes + 50 tonnes). So, it is a problem of unbalanced transportation. The total supply is larger than thesurplus supply available. We can prepare the initial feasible table as follows by lowest cost entry method:

Initial feasible table by lowest cost entry method

Warehouse Factory R S T X

Availability

30 10 50 0 60 40 A 10 10 50 -10

100

20 40 60 0 70 120 B 20 20 60 4

200

Required 70 60 50 (120) 300


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In the above table it can be observed that m + n- 1 =5. It means 5 cells are occupied and the rule m + n - I issatisfied. Hence the solution is feasible.

In the above table, all implicit costs are either less or equal to the actual transportation cost hence thisrepresents the optimal solution.

The least cost transportation schedule will be as follows:

Rs.From factory A to warehouse S : 40 tonnes @ 10 per tonne = 4,00From factory A to warehouse T : 40 tonnes @ 50 per tonne = 2,000From factory B to warehouse R : 70 tonnes @ 20 per tonne = 1,400From factory B to warehouse T : 10 tonnes @ 60 per tonne = 600Minimum total transportation cost = 4,400

4.2 Transportation

Transportation applications relate to a LPP where goods are to be transported from “m” production locations(factories) to “n” sales locations (warehouses). The objectives are (a) To meet the differing availabilitiesand requirements of these locations and (b) To minimise the total transportation costs.

The transportation application can be solved in three stages:

(a) Preliminary Check,

(b) Initial Basic Feasible Solution (IBFS),

(c) Optimality Test.

Methods of finding the initial basic feasible solution to a transportation problem:

IBFS can be determined using any of the following methods:

(a) Northwest Corner Rule,

(b) Least Cost Cell Method,

(c) Vogel’s Approximation Method (VAM).

Sages involved in determining the solution to the Transportation Problem.

(1) Stage 1: Preliminary check involves the following:

(a) Verify Objective = Minimisation. In case of profit matrix, convert the same into an OpportunityLoss Matrix, by subtracting each number from the highest number in the matrix.

(b) Verify Nature of data = Balanced.

Data is said to be balanced if Total Availability = Total Requirement.

In case of unbalanced data, a dummy column or row should be introduced with zero transportation costs.

(2) Stage 2: IBFS can be determined using any of the following methods:

(a) Northwest Corner Rule,

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(b) Least Cost Cell Method,

(c) Vogel’s Approximation Method (VAM).

(3) Stage 3: Optimality Test: It consists of the following steps

(a) Computation of margin numbers, ‘U’ and ‘V for all rows and columns such that U

+ V = Cost of allocated cells,

(b) Computation of U + V for unallocated cells,

(c) Computation of Cost Less (U + V) for unallocated cells, i.e. Step 1 minus Step 2 above.

Northwest Corner Rule in determining the IBFS.Step Procedures

1 � Ensure Availability = Requirement, by inserting dummy row or column, if required.

2 � Go to top left hand corner cell of the matrix � Compare availability and requirement. � Allocate availability or requirement, whichever is less, to that cell. � Cancel the row or column where availability or requirement is exhausted.

3 � Go to top left hand corner cell of the resultant matrix (after cancellation of row or column in Step 2)

� Repeat Step 2 procedure till all the row availability and column requirements are satisfied.

Areas of Application:

Northwest Corner Rule may be used in transportation applications

• Within the campus of an organisation as cost differences are not significant.

• Obligations where cost is not the criteria for decision-making.

Least Cost Cell Method of determining the IBFS.

Advantage: The cost associated with each route is taken into consideration. So, this method leads to abetter allocation than North West Corner Rule Method.

Step Procedures 1 • Ensure Availability = Requirement, by inserting dummy row or column, if

required. 2 • Identify the cell with the lowest cost. In case of a tie, arbitrary selection may

be made. • Compare availability and requirement. • Allocate availability or requirement, whichever is less, to that cell. • Cancel the row or column where availability or requirement is exhausted.

3 • Identify the next lowest cell in the matrix. • Repeat Step 2 procedure till all the row availability and column requirements

are satisfied.


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Vogel’s Approximation Method (VAM) for obtaining IBFS

Steps involved in Optimality Test:

Optimality test involves the following steps:

(1) Table 1: Ui + Vj for allocated cells:

(a) Select the row/column with maximum number of allocations.

(b) For that row/column, Ui/Vj is equal to zero. [Ui for Rows; Vj for Columns]

(c) The other set of numbers Ui/Vj are computed in such a way that Ui + Vj = Cost of allocated cells.

Note : The Ui + Vj table can be completed only if the IBFS is non-degenerate.

IBFS is said to be degenerate if number of allocations < (No. of rows + No. of Columns - 1).

In case of degeneracy, a dummy allocation “e” (a number very close to zero) is made in the least costunallocated cell falling in a non-dummy row or column.

(2) Table 2: Ui + Vj for unallocated cells:

(a) Draw a matrix for the given rows and columns.

(b) Block out the allocated cells.

(c) Compute Ui + Vj (total of margin numbers) for all unallocated cells.

(3) Table 3: Net Evaluation Table = Δij

(a) Draw a matrix for the given rows and columns.

(b) Block out the allocated cells.

(c) Compute Cost Difference for all unallocated cells. Cost Difference = Number in Table (1) LessNumber in Table (2).

Decision:

(a) If all numbers in the Net Evaluation Table are non-negative (i.e. > 0), IBFS is optimal and unique.

Steps Procedures 1 � Compute Cost Differences for each row and column.

� Cost Difference is the difference between the least cost and the next least cost in that row/column.

� In case of tie in least cost, Cost Difference = 0. 2 Ascertain the maximum of cost differences and select that row or column for

allocation 3 Choose the least cost cell in the selected row or column for allocation. 4 � Compare availability and requirement for that cell.

� Allocate availability or requirement, whichever is less, to that cell. � Cancel the row or column where availability or requirement is exhausted.

5 Compute Cost Differences for the resultant matrix and repeat the above procedure till all row availability and column requirements are satisfied.

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(b) If all numbers are positive and one of the cells contains zero, IBFS is optimal but not unique. Alternativesolution exists. Total number of solutions = Number of Zeroes + 1

(c) If any one number in Table 3 is negative, solution is not optimal. Reallocation should be made todetermine Alternative Basic Feasible Solution (ABFS).

(d) ABFS is tested for optimality by adopting the same procedure.

Concept of “loop” diagram in Transportation

(a) Identify the worst negative in Table 3. If there is a tie, choose the least cost cell.

(b) Draw a loop with the following principles:

• Loop should commence from and end in the selected worst negative cell.

• It should have only allocated cells as its other corners.

• Only horizontal and vertical lines (not diagonal) shall be permitted.

• Loop should result in a even sided figure.

(c) Identify the selected cell as having scope for allocation. It is marked with plus (+) sign.

(d) Other corners of the loop are identified with (-) and (+) signs alternatively.

(e) ABFS is determined by reference to reallocation as specified by the corners of the loop.

(f) Reallocation is done as under

• Identify the allocations to the negative cells (corners marked with (-) sign),

• Select the minimum out of the above allocations,

• Add this minimum to (+) corners and subtract this minimum from (-) corners,

Other allocations remain unaffected.

(g) This ABFS is tested for optimality using the procedure outlined in the earlier question. In case negativesstill arise in the Net Evaluation Table, the reallocation procedure should be continued further.

Treatment of different situations in the context of Transportation Problems:

Situation Treatment Total availability not equal to total requirement

• It is called unbalanced transportation application. • Introduce dummy row or column to balance availability and

requirement. • All entries (costs) of the dummy row/column shall be zeroes.

Maximisation objective

• Select the highest element of profit in the matrix. • Subtract each element from the maximum element. • Take the resultant opportunity cost matrix with minimisation

objective, for allocation procedure.


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Problem 1.A Company has four factories Fl, F2, F3 and F4 manufacturing the same product. Production and rawmaterial costs differ from factory to factory and are given in the following table. The transportation costsfrom the factories to sales depots S1, S2 and S3 are also given. The sales price and the total requirement ateach depot as also the product capacity at each factory is also stated. Determine the most profitableproduction and distribution schedule and the corresponding profit. The surplus production should betaken to yield zero profit.

Degeneracy • Optimality test can be done only if number of allocations = [No. of rows + No. of Columns - 1]

• If number of allocations is less, all Ui and Vj numbers cannot be computed. This situation is called degeneracy.

• If IBFS is degenerate, introduce a dummy allocation (e), a small number very close to zero, in the least cost unallocated cell falling in a regular row or column. (i.e. non-dummy row or column)

Existence of zeroes in he Net Evaluation Table

• Zero in the Net Evaluation Table (Table 3) indicates the availability of alternative solutions. (provided all other numbers are positive)

• The alternative optimal solutions can be determined by drawing a loop commencing from such cell having zero as net evaluation.

• Reallocation will be based on the loop drawing procedure outlined above.

Prohibited routes • Sometimes some routes may not be available due to reasons like bad road conditions, strike etc.

• Such prohibited routes are identified with a high cost. represented by or “M”, a very large cost close to infinity.

• Due to assignment of very large cost, such routes would automatically be eliminated in the final solution.

Situation Treatment

Particulars Fl F2 F3 F4 Sales price per unit

Requirements per unit

Production Cost per unit 15

18

14 13

Raw material Cost per unit 10 9 12 9 Transportation cost per unit to

S1 3 9 5 4 34 80 S2 1 7 4 5 32 120 S3 5 8 3 6 31 150

Production capacity (units) 10 150 50 100

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Solution:From the data given, the profit matrix is constructed as follows:Profit = Selling Price - Production Cost - Material Cost - Transportation Cost

• In the above matrix, the objective is maximization of profit. This is converted into a opportunity costminimization objective by subtracting each element from the highest element i.e. 8.

• Also, the above matrix is unbalanced data since total Capacity is 310 units and total Requirement is350 units. This is converted into balanced data by introducing the dummy factory F- 5 with zero as itsentries.

• The revised matrix (balanced minimization) is given below.

Place F -1 F - 2 F - 3 F - 4 Requirement

S - 1 6 -2 3 8 80

S - 2 6 -2 2 5 120

S - 3 1 - 4 2 3 150 Capacity 10 150 50 100

Place F -1 F - 2 F - 3 F - 4 F - 5 Requirement

S - 1 2 10 5 0 0 80

S - 2 2 10 6 3 0 120

S -3 7 12 6 5 0 150 Capacity 10 150 50 100 40 350

The above matrix is now amenable for applying VAM.

Initial Basic Feasible Solution (IBFS) is determined as under:

Cost Diff:

Note: Upon full allocation, the relevant row/column may be cancelled in pencil.

I 0 0 1 3 0 In the above IBFS, II 0 0 1 3 _ • Number of allocated cells is 7.

III 5

2 0 2 − • m + n − 1 (i.e. Rows + Columns −1)= 3+ 5-1=7.

IV _ 2 0 2 _ V _ 2 0 _ _ VI − 2 _ − −

Hence, there is no degeneracy. This can be tested for optimality.

Place F-1 F-2 F-3 F-4 F-5 Requirement Cost Differences I II III IV V VI

80 80/0 0 2 — — — — S−1 2 10 5 0 0

10 90 20 120/110/90/0 2 1 1 3 4 10 S−2

2 10 6 3 0 60 50 40 150/110/60 5 1 1 1 6 12

S−3 7 12 6 5 0

Capacity 10/0 150/60/0 50/0 100/20/0 40/0 350


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OPTIMALITY TEST:

Table 1 = Ui + Vj for allocated cells computed as below:

Ui + Vj 2 − 0 = 2 10 − 0 = 10 6 − 2 = 4 3 − 0 = 3 0 − 2 = −2 80

0 −3 = −3 2 10 5 0 0 10 90 20

(base) = 0 2 10 6 3 0 60 50 40 12 – 10 = 2 7 12 6 5 0

Table 2 = Ui + Vj for unallocated cells computed as below:

−3+2 = −1 −3+10 = 7 −3+ 4 = 1 −3+(−2) = −5 0 + 4 = 4 0 + (−2)= −2

2+2 = 4 2+3 = 5

Table 3 = Net Evaluation Table (NET) = Table 1 - Table 2 for unallocated cells is computed below:

2− (−1) = 3 10 −7= 3 5 − 1= 4 0 − (−5) = 5 6 − 4 = 2 0 − (−2)= 2

7 − 4 = 3 5 − 5 = 0

Conclusion: Since all entries in NET are non-negative, the IBFS is optimal. Since one entry in NET is zero,the above optimal solution is not unique. One alternative solution exists.

Computation of Maximum Profit: (from the profit matrix formulated first)

Place F − l F −2 F −3 F−4 F−5 S- l 80 × 8 = 640 S-2 10×6 = 60 90×(−2) = −180 20 × 5 = 100 S-3 60×(−4) = −240 50 × 2 = 100 40 × 0 = 0

Determination of Alternative Optimal Solution: Since alternative solution is available, (one entry in NET= 0), the alternative optimal solution is determined by drawing a loop from the “Zero” entry in the NET.The loop is shown below in the NET of the IBFS.

80 2 10 5 0 0 10 90 + 20=110 20 2 10 6 3 0 60 – 20 = 40 50 40 7 12 6 5 0

2− (−1) = 3 10 −7= 3 5 − 1= 4 0 − (−5) = 5 + ve 6 − 4 = 2 − ve 20 0 − (−2)= 2

7 − 4 = 3 -ve 60 5 − 5 = 0 + ve

Since the least allocation of the negative corners of the loop is 20, the alternative solution (optimal solution)determined by adding 20 to the positive corners, subtracting 20 from the negative corners and leaving theother cells undisturbed.

The alternative allocation is shown below:

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Resource Management

The profit form the above alternative allocation is:

Place F − 1 F − 2 F − 3 F − 4 F − 5 S−1 80 × 8 = 640 S−2 10 × 6 = 60 110 × (−2) = – 220 S−3 40 × (−4) = −160 50 × 2 = 100 20 × 3 = 60 40 × 0 = 0

Maximum Profit = Total of above = Rs.480

Note: The alternative solution need not be tested further for optimality, since the IBFS is already optimal.

Problem 2.

A company has three factories and four customers. It furnishes the following schedule of profit per unit ontransportation of goods to customers in rupees. You are required to solve the transportation problem tomaximize the profit. Determine the resultant optimal profit.

Factory / Customer

A B C D E Supply

P 4 19 22 11 0 100 Q 0 9 14 14 0 30 R 6 6 16 14 0 70

Demand 40 20 60 30 50

Initial Basic Feasible Solution is determined as under:

Factory / Customer

A B C D E Supply Cost Differences

I II III IV 10 60 30 4 7 7 7

A 4 19 22 11 0 100/90/0 30 0 9 − − B 0 9 14 14 0 30/0 20 50 6 0 0 − C 6 6 16 14 0 70/20/0

Demand 40/10/0 20/0 60/0 30/0 50/0 200

I 4 3 2 3 0 In the above IBFS, II 4 3 2 3 − • Number of allocated cells is 6. III 2 13 6 3 − • m + n- 1 (i.e. Rows + Columns- 1) IV 4 − 2 11 − = 3 + 5-1 = 7.

Hence, there is a degeneracy.

Cost Diff:

Factory / Customer

A B C D Supply

P 40 25 22 33 100 Q 44 35 30 30 30 R 38 38 28 30 70

Demand 40 20 60 30

Solution: The above problem is profit maximization. Hence it should be converted into an opportunityloss matrix by subtracting the highest element i.e. 44 from the rest. Also the problem is unbalanced asdemand = 150 and Supply = 200. Hence it should be balanced by introducing a dummy column with allentries as 0. Performing the above operations, the resultant balanced minimization matrix is -


259

To overcome degeneracy, a dummy allocation represented by ‘e’ (where ‘e’ = 0.00000.......1) is made in theleast cost unallocated cell falling in a regular row or column i.e. non dummy row or column. [It should benoted that degeneracy does not make the IBFS as wrong answer; it is only a situation not a defect.]

OPTIMALITY TEST:


+ ve 19+4 = 15 − ve 60 0−(−2) = 2 9 + 0 = 9 14 − 18 = −4 14 − 7 = 7 0−(−6) = 6 − ve ‘e’ 16−24 = −8 14 − 13 = 1

Ui & Vj 4 −0 = 4 6 − 2 = 4 22 − 0 = 22 11 − 0= 11 0 − 2 = − 2

(base) 0 10 60 30 4 19 22 11 0

0 − 4= −4 30 0 9 14 14 0

6 − 4 = 2 e 20 50 6 6 16 14 0

0 + 4 = 4 0+(−2) = −2 − 4 + 4 = 0 − 4 + 22 = 18 − 4 + 11=7 − 4+ (−2) = − 6 2 + 22 =24 2 + 11 = 13

19−4 = 15 0−(−2) = 2 9−0 = 9 14 − 18 = − 4 14 − 7 =7 0−(−6) = 6 16 − 24 = − 8 14 −13 =1


Table 3 = Net Evaluation Table (NET) = Table 1- Table 2 for unallocated cells is Table computed below:

There are two negative elements in the NET, hence allocation is not optimal. The loop is created as below-

ABFS 1: The new Ui + Vj for allocated cells is computed from the above.


Ui & Vj 4 −0 = 4 6−(−6) =12 22− 0 = 22 11− 0 =11 0 − (−6) = 6 (base) 0 10 60 30

4 19 22 11 0 0 − 4 = − 4 30

0 9 14 14 0 16− 22 = − 6 20 e 50

6 6 16 14 0

‘e’ is subtracted / added.

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Resource Management


19−12 = 7 0 − 6 = −6 9−8= 1 14−18 =−4 14−7 =7 0 − 2 = −2

6 −(−2) = 8 14−5 =9

Ui & Vj 4 − 0 = 4 6 − (−6) =12 22 − 0 = 22 11− 0 =11 0 − (0) = 0 (base) 0 10 60 30 50

4 19 22 11 0 0 − 4 = −4 30 0 9 14 14 0

20 50 16−22= −6 6 6 16 14 0

19 − 12 = 7 9 −8 = 1 14 − 18 = −4 14 −7 = 7 0 − (−4) = 4

6 −(−2) = 8 14 −5 = 9 0 − (−6) = 6

0 + 12 = 12 − 4 + 12 = 8 −4 + 22 = 18 −4 + 11 = 7 − 4 + 0 = −4

− 6 + 4 = −2 −6 + 11 =5 − 6 + 0 = − 6

Table 3 = Net Evaluation Table (NET) = Table 1 - Table 2 for unallocated cells is computed below -

There are negative elements in the NET; hence allocation ABFS 1 is not optimal. The loop is created asbelow -

ABFS 2 : The new Ui + Vj for allocated cells is computed from the above.




There are negative elements in the NET, hence allocation is not optimal. The loop is created as below:

+ ve 19 − 12 = 7 − ve 30 9 −8 = 1 14 − 18 = −4 14 −7 = 7 0 − (−4) = 4

6 − (−2) = 8 14 −5 = 9 0 − (−6) = 6

0+ 12= 12 0 + 6 = 6 − 4+12 = 8 −4 + 22 = 18 −4+11=7 −4 + 6 = 2

−6 + 4 = −2 −6+11=5

19−12 = 7 − ve 60 0 − 6 = −6 9−8= 1 14−18 = −4 14−7 =7 0 − 2 = −2

6 −(−2) = 8 + ve 14−5 =9 − ve 50


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Ui & Vj 4 − 0 = 4 6− (−2) = 8 14 (−4) = 18 11− 0 =11 0 − 0 = 0 (base) 0 20 30 50

4 19 22 11 0 0 − 4 = − 4 20 10

0 9 14 14 0 16 − 18 = -2 20 50

6 6 16 14 0


Table 3 = Net Evaluation Table (NET) = Table 1- Table 2 for unallocated cells is computed below.

Since there are no non-negative elements in the NET, ABFS 3 is optimal and unique. Computation ofMaximum Profit: (from the profit matrix formulated first)

Maximum Profit = Total of above = Rs.5130

Problem 3.

The information on the available supply to each warehouse, requirement of each market and the unittransportation cost from each warehouse to each market is given below:

0 + 8 = 8 0+18 = 18 − 4 + 8 = 4 − 4 + 11 = 7 − 4 + 0 = − 4

− 2 + 4 = 2 − 2 + 11 = 9 − 2 + 0 = − 2

19−8 = 11 22−18 = 4 9−4 = 5 14−7 =7 0 − (−4) = 4

6−2 = 4 14−9 = 5 0 − (−2) = 2

Place A B C D E P 20 × 40 = 800 30×33=990 50×0 = 0 Q 20 × 44 = 880 10×30 = 300 R 20 × 38 = 760 50×28= 1400

Market Supply Warehouse

Ml M2 M3 M4 A 5 2 4 3 22 B 4 8 1 6 15 C 4 6 7 5 8

Demand 7 12 17 9

The shipping clerk has worked out the following schedule from his experience

Units 12 1 9 15 7 1 From warehouse A A A B C C To – market M2 M3 M4 M3 Ml M3

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Resource Management

You are required to-

(i) Check and see if the clerk has the optimal schedule

(ii) Find the optimal schedule and minimum total shipping cost and

If the clerk is approached by a carrier of route C to M2, who offers to reduce his rate in the hope of gettingsome business, by how much should the rate be reduced before the clerk considers giving him an order.

Solution:

Initial Basic Feasible Solution is determined as under from the data given above

Place M-l M-2 M-3 M-4 Requirement Cost Differences I II III IV V VI

12 1 9 A 5 2 4 3

15 B 4 8 1 6

7 1 C 4 6 7 5

Demand

I In the above IBFS, II • Number of allocated cells is 6. III • m + n-1 (i.e. Rows + Columns-1) IV = 3 + 4-1=6. V VI


Cost Diff :

Cost Differences have not been computed since the Clerk’s allocation is taken as the IBFS.

OPTIMALITY TEST:


Ui & Vj 4 − 7 = − 3 2 − 4= −2 0(base) 3 − 4= −1 4 − 0 = 4 12 1 9

5 2 4 3 1 − 0= 1 15

4 8 1 6 7 −0 = 7 7 1

4 6 7 5

4 + (−3) = 1 1 + (−3) = −2 l + (−2) = −l l + (−l) = 0

7 + (−2) = 5 7 + (−1) = 6



263

Table 3 = Net Evaluation Table (NET) = Table1-Table 2 for unallocated cells is computed below:

5 − 1=4 4 −(−2) = 6 8 − (−l) = 9 6−0 = 6

6 − 5 = 1 5−6 = −1

5 − 1 = 4 + ve −ve 4 − (−2) = 6 8 −(−l) = 9 6 − 0 = 6

6 − 5 = 1 −ve + ve 5 − 6 = −l

Ui & Vj 4−2 = 2 2−0 = 2 4−0 = 4 3−0 = 3 (base) 0 12 1+1 = 2 9−1=8

5 2 4 3 1 − 4 = −3 15

4 8 1 6 5 − 3 = 2 7 0+1=1

4 6 7 5

There is one negative element in the NET, hence scheduling by the clerk is not optimal. 1 is the leastallocated to the negative corner. The loop is created as below -




0 + 2 = 2 − 3+2 = −1 −3 + 2 = −1 −3 + 3 = 0

2 + 2 = 4 2 + 4 = 6

5 − 2 = 3 4 − (−l) = 5 8 − (−l) = 9 6 − 0 = 6

6 − 4 = 2 7 − 6=1

Place M−l M−2 M−3 M−4 A 12×2 = 24 2×4 = 8 8 × 3 = 24 B 15× 1 = 15 C 7×4 = 28 1 ×5 = 5


All elements in the NET are non-negative; hence the revised allocation is optimal and unique.

The optimal allocation and cost is -

Minimum Cost - Total of above = Rs.104

If the clerk wants to consider the carrier of route C to M - 2 for giving an order, then this transportation costshould be less than the carrier of routes C to M - 1 and C to M - 4 (which are presently allocated cells /routes.) Hence the carrier of route C to M - 4 has to bring down his rate to Rs.3 from the present Rs.6.

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Resource Management

Problem 4.

X Company is interested in taking loans from banks for its projects — P, Q, R, S, T. The rates of interest andthe lending capacity differ from bank to bank. All these projects are to be completed. The relevant detailsare provided below. Assuming the role of a consultant, advice the Company as to how it should take theloans so that the total interest payable is least. Find out alternate optimum solutions, if any.

Interest rate in % for projects Max CreditSource Bank

P Q R S T (in 000s) Private Bank 20 18 18 17 17 Any

amount Nationalised Bank 16 16 16 15 16 400 Co-operative Bank 15 15 15 13 14 250 Amount required (in 000s)

200 150 200 125 75

Solution.

Total amount required as loan = Rs.750 (000s). The private bank can give any amount. The data ismade balanced by putting 100 against the private bank.

Initial Basic Feasible Solution is determined as under:

P Q R S T Amount Cost Differences Part.

I II III IV V VI 100 Private 20 18 18 17 17 100/0 0 1 0 0 0 18

150 50 200 National 16 16 16 15 16 400/250/50/

0 1 0 0 0 0 16

50 125 75 Co-op 15 15 15 13 14 250/125/50/

0 1 1 0 − − −

Requ-ired

200/150/0

150/0 200/0 125/0 75/0 750

Cost Diff:

I 1 1 1 2 2 In the above IBFS, II 1 1 1 − 2 • Number of allocated cells is 7. III 1 1 1 _ − • m + n - 1 (i.e. Rows + Columns - 1) IV 4 2 2 − − = 3 + 5−1=7. V − 2 2 − − VI − 2 − − −



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Optimality Test:


3 + 15 = 18 3 + 15 = 18 3 + 13 = 16 3+ 14=17 1 + 13 = 14 1 + 14= 15 0 + 15 = 15 0 + 15=15


Ui & Vj 15−0=15 16−1 = 15 16−1 = 15 13−0= 13 14−0=14 18−15=3 100

20 18 18 17 17 16−15 = 1 150 50 200

16 16 16 15 16 (base) 0 50 125 75

15 15 15 13 14


Solution is optimal since all elements in NET are non-negative. However there are four zeroes and so thesolution is not unique. There are four alternate solutions.

Computation of Minimum Cost: (amounts in ‘000s and interest rate in %)

20−18 = 2 18−18 = 0 17−16= 1 17−17 = 0 15−14= 1 16−15 = 1 15−15 = 0 15−15 = 0

Particulars P Q R S T Private 100× 18= 18 National 150×16 =24.00 50× 16 = 8 200×16= 32 Co-op 50×15 = 7.50 125×13= 16.25 75×14= 10.50

20 − 18 = 2 − ve 100 18 − 18 = 0 17 − 16 = 1 17 − 17 = 0 + ve − ve 15 − 14 = 1 16 − 15 = 1

15 − 15 = 0 15 − 15 = 0

Minimum Cost = Total of above = Rs.l,16,250

Determination of Alternative Optimal Solution -1:

The alternative optimal solution is determined by drawing a loop from the “Zero” entry in the NET. Theloop is shown below in the NET of the IBFS.

Since the least allocation of the negative corners of the loop is 100, the alternative solution (optimal solution)is determined by adding 100 to the positive corners, subtracting 100 from the negative corners and leavingthe other cells undisturbed.

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Resource Management


Minimum Cost = Total of above = Rs.1,16,250

Determination of Alternative Optimal Solution 3: The alternative optimal solution is determined bydrawing a loop from the “Zero” entry in the NET. The loop is shown below in the NET of the IBFS.



Minimum Cost = Total of above = Rs.l, 16,250

Determination of Alternative Optimal Solution - 3: The alternative optimal solution is determined bydrawing a loop from the “Zero” entry in the NET. The loop is shown below in the NET of the IBFS

100 – 100 = 0 0 + 100 = 10020 18 18 17 17

150 50 + 100 = 150 200 - 100 = 10016 16 16 15 16

50 125 7515 15 15 13 14

Particulars P Q R S T

Private 100 x 18 =18.00

National 150 x 16 = 24.00 150 x 16 = 24.00 200 x 16=32.00

Co- op 50 x 15 = 7.50 125 x 13 = 16.25 75 x 14 = 10.50

The cost from above alternative re-allocation is :

17 – 16 = 1 17 – 17 = 015 – 14 = 1 16 – 15 = 1

20 – 18 = 2 18 – 18 = 0

15 – 15 = 0 15 – 15 = 0

+ ve

20150 + 50 = 200

1650 – 50 = 0

18

16 16 1615

18 17 17100

50–50=0

0 + 50 = 50

200

12515 15 15 13 14

ParticularsPrivateNationalCo-op

P Q R S T100 x 18 = 18.00

200 x 16 = 32.0050 x 15 = 7.50

200 x 16 = 32.00125 x 13 = 16.25 75 x 14 = 10.50

The cost from the above alternative re-allocation is :

75

+ ve

-- ve50


267


Minimum Cost = Total of above = Rs.l,16,250

Determination of Alternative Optimal Solution - 4: ]

The alternative optimal solution is determined by drawing a loop from the “Zero” entry in the NET. Theloop is shown below in the NET of the IBFS.


Minimum Cost = Total of above = Rs. 1,16,250

20 – 18 = 2 18 – 18 = 0 17 – 16 = 1 17 – 17 = 015 – 14 = 1 16 – 15 = 1

15 – 15 = 0 15 – 15 = 0+ve– ve50

– ve200

Particulars P Q R S T

Private 100 x 18 = 18.00

National 200 x 16 = 32.00 50 x 16 = 8.00 150 x 16=24.00

Co- op 50 x 15 = 7.50 125 x 13 = 16.25 75 x 14 = 10.50

20150 + 50 = 200

1650 – 50 = 0

18

16 16 1615

18 17 17100

50 200 – 50 = 150

12515 15 15 13 14

750 + 50 = 50

The cost from above alternative re-allocation is :

17 – 16 = 1 17 – 17 = 015 – 14 = 1 16 – 15 = 1

20 – 18 =2 18 – 18 = 0

15 – 15 = 0 15 – 15 = 0+ ve– ve 150

The cost from the above alternative re-allocation is :

ParticularsPrivateNationalCo-op

P Q R S T 25 x 18 = 4.50

75 x 16 = 12.00 125 x 16 = 20.00 200 x 16 = 32.00125 x 13 = 16.25

75 x 17 = 12.75

20150 - 75 = 75

1650 + 75 = 125

18

16 16 1615

18 17 17100 - 75 = 25

200

12515 15 15 13 14

50 + 75 = 125

- ve = 75

– ve 100+ ve

0 + 75 = 75

75 - 75 = 0

125 x 15 = 18.75

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Resource Management

4.3 Replacement of Machine and other Relevant Concept

Replacement of Machines and Equipment

Machines are purchased and replacement of old machines-are made mainly for two reasons:

1. To increase the productive capacity and

2. To reduce cost of production.

Various other reasons for replacement are the following,

1. To get rid of worn out, broken down or obsolete machines,

2. To accommodate larger sizes of work and increase the machine capacity,

3. To reduce labour cost by introducing semi-automatic machines or machines more than one of whichcan be operated by a single operator,

4. To simplify operations by using automatic machines which are capable of performing variety of workusually performed by a number of different machines,

5. To minimise repair cost and reduce idle time.

An analysis of the above six reasons will lead to either increase in capacity or reduction in cost or both.

Factors on which equipment is replaced: The replacement plan depends on evaluation of present andreplacement machines from the point of view of technical suitability and cost saving features.

The points to check for replacement studies vary from industry, to industry on management conditionsand management policies. But some factors are common to practically all cases. These are:

Technical Factors:

(i) Inadequacy from the stand paint of range, speed, accuracy, strength, rigidity, output and capacity,

(ii) Obsolescence and equipment worn out condition,

(iii) Special advantage of the new machine as to easiness of set ups convenience of operation, safety,reliability performance, control panels and special features,

(iv) Flexibility and versatility of the machine.

Cost Factors:

(i) High repair cost of existing machine,

(ii) High remodelling cost of existing machine,

(iii) Less chance of spoilage and rejection work causing: saving in cost,

(iv) Faster rate of production causes lower cost,

(v) Replacement of skilled -workers by semi-skilled and: unskilled workers leading to lesser labour cost,

(vi) Compactness of the machine leading to a saving in-space which means saving of overhead costs,

(vii) Machine pay back period i.e. how soon the cost of the equipment is recovered,

(viii)Life of the new machine giving effective service,


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(ix) Flexibility and versatility of the machine tending to reduce idle time cost with changes in methods ofproduction-which might occur in future,

(x) Availability of funds for the acquisition of the equipment or possibility of special arrangement likehire-purchase or government loans or other accommodations.

Replacement Programmes: It is prudent to have phased .policies of machine replacement plans than towait until breakdowns occur causing production hold ups. There are different forms of the programme.

1. A definite amount of money or a certain percentage of earning of the company is used each year toreplace existing machines which are either superseded by improved models or are not in tip topcondition or are having insufficient capacity.

2. Replacement is made of the oldest or most inadequate machine each year by upto date machines ofgreater accuracy or higher capacity. Some times automatic machines are gradually introduced in thisway which is capable of doing several operations with lesser number of operators.

3. The economy of working on various machines are studied .and replacement of machines are madeonly to have a definite cost reduction.

Whatever may be the programme, replacement question is to be carefully considered in prosperous andnormal years only. In slum or dull periods, replacement should not be done unless it is unavoidable.

The basic concept of ‘Replacement Theory’ is to take decision about the time when an item of ‘CapitalAsset’ should be replaced by another of the same type or by a different one. In other words ‘ReplacementTheory’ concerns about ‘optimum period of replacement’.

For the purpose of ‘Replacement Theory’ Capital equipment are basically divided under two broadcategories:

1. Replacement policy for Equipment which detoriates with time gradually;

2. Items which fail suddenly and cannot be made workable by incurring repairing costs.

Again, replacement of Capital equipment which gradually detoriates with time, can be worked-out undertwo different ways:

(i) Ignoring the concept of time value of money and considering the time value ofmoney.

(ii) When time value of money is not considered: Determination of optimum period of‘Replacement’ Let us, consider the following formula:

T(r) = C - S + ∑=

r

1t

Mt

where, T(r) denotes owning and maintaining cost of keeping an equipment for ‘r years’

C = Capital cost of the equipment

S = Salvage value of the equipment at the end of ‘r years’

Mt = Cost of maintenance in year t

Accordingly, the average cost, A(r) can be calculated as under:

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Resource Management

A(r) = r

1 ⎥

⎦

⎤⎢⎣

⎡ +− ∑=

r

T

MtSC1

Decision: When to replace? The optimal replacement period would be the year, in which A(r) = Averagecost is minimum.

(ii) When time value of money is considered: In this case the concept of present value of money isconsidered. Considering the interest rate, which is also termed as discounting factor or discountingrate with that the present values of maintenance costs are to be evaluated for each year and finalannualised cost, for different years are also evaluated. Decision: When to replace? The decision willbe taken by considering the following rules:

A. Replacement will not be made if annualised cost > next period’s cost.

B. Replacement will be made in that year only, when annualised cost < next period’s cost. In otherwords, replacement will be made every r years.

when A(r) = the annualized cost of replacing every r years is the minimum.

Important Assumptions

(a) Salvage/scrap value of the equipment to be replaced should be considered as ‘nil’

(b) The maintenance costs are assumed to be incurred at the beginning of the year.

Replacement of items which fail suddenly

Sometimes, the failure of an item may cause a complete breakdown of the system. The costs of failure, insuch a case will be quite higher than the cost of the item itself. So, it is important to know in advance as towhen the failure is likely to take place so that item can be replaced before it actually fails and the time offailure can be predicted from the probability distribution of failure time obtained from past experience andthen, to find the optimal value of time t which minimises the total cost involved in the system. Mainly twotypes of replacement policies are there:

(1) Individual replacement policy in which an item is replaced immediately after it fails.

(2) Group replacement policy in which all items are replaced, irrespective of whether they have failed ornot, with a provision that if any item fails before the optimal time, it may be individually replaced.

Let us, explain through example how ‘Replacement of items which fail suddenly’ is considered: Followingfailure rates have been observed for a certain type of light bulbs:

Week: 1 2 3 4 5 Per cent failing by the end of week: 10 25 50 80 100

There are 1,000 bulbs in use, and it costs Rs. 2 to replace an individual bulb which has burnt out. If all bulbswere replaced simultaneously it would cost 50 paisa per bulb. It is proposed to replace all bulbs at fixedintervals of time, whether or not they have burnt out, and to continue replacing burnt out bulbs as andwhen they fail. At what intervals all the bulbs should be replaced? At what group replacement price perbulb would a policy of strictly individual replacement become preferable to the adopted policy?

Solution: Let, Pi be the probability that a light bulb, which was new when placed in position for use, failsduring the ith week of its life.


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Thus, following frequency distribution is obtained assuming to replace burnt out bulbs as and when theyfail,

P1 = the prob. of failure in Ist week = (10/100) = 0.10P2 = the prob. of failure in IInd week = (25 - 10/100 = 0.15P3 = the prob. of failure in IIIrd week = (50 - 25)/l00 = 0.25P4 = the prob. of failure in IVth week = (80 - 50)/100 = 0.30P5 = the prob. of failure in Vth week = (100 - 80)/100 = 0.20Total = 1.00

Since the sum of all probabilities can never be greater than unity, therefore all further probabilities P6, P7

and P8 so on, will be zero. Thus, a bulb that has already lasted four weeks is sure to fail during the fifthweek. Furthermore, assume that

(i) Bulbs that fail during a week are replaced just before the end of that week, and

(ii) The actual percentage of failures during a week for a subpopulation of bulbs with the same age is thesame as the expected percentage of failures during the week for that subpopulation.

Let Ni be the number of replacements made at the end of the ith week, if all 1000 bulbs are new initially.Thus,

N0 = N0 = 1,000 N1 = N0pl = 1,000 × 0.10 = 100 N2 = N0p2 + N1p1 = 1,000×0.15+ 100×0.10 = 160 N3 = N0P3 + N1p2 + N2p1 = 1,000×0.25+ 100×0.15 + 160×0.10 = 281 N4 = N0p4 + N1p3 + N2P2 + N3p1 = 377 N5 = N0p5 + N1p4 + N2P3 + N3p2 + N4p1 = 350 N6 = 0 + N1p5 + N2p4 + N3p3 + N4p2 + N5p1 = 230 N7 = 0 + 0 + N2p5 + N3p4 + N4p3 + N5p2 + N6p1 = 286

and so on.

It has been observed that the expected number of bulbs burnt out in each week increases until 4th week andthen decreases until 6th week and again starts increasing. Thus, the number will continue to oscillate andultimately the system settles down to a steady state in which the proportion of bulbs failing in each weekis the reciprocal of their average life.

Since the mean age of bulbs

= l×p1+2×p2 + 3×p3 + 4×p4 + 5 ×p5

= 1 × 0.10 + 2 × 0.15 + 3 × 0.25 + 4 × 0.30 + 5 × 0.20 = 3.35 weeks.

the number of failures in each week in steady state becomes

= 1,000/3.35 = 299 and the cost of replacing bulbs only on failure = 2 × 299 (at the rate of Rs. 2 per bulb) =Rs. 598.

Since the replacement of all 1,000 bulbs simultaneously costs 50 paise per bulb and replacement of anindividual bulb on failure costs Rs. 2, therefore cost of replacement of all bulbs simultaneously is as givenin the following table:

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Resource Management

The cost of individual replacement in the third week exceeds the average cost for two weeks.

Thus it would be optimal to replace all the bulbs after every two weeks; otherwise the average cost willstart increasing.

Further, since the group replacement after one week costs Rs. 700 and the individual replacement after oneweek costs Rs. 598, the individual replacement will be preferable.

Replacement—Staffing Problem

Problems concerning recruitment and promotion of staff can sometimes be analysed in a manner similarto that used in replacement problems in industry.

A faculty in a college is planned to rise to strength of 50 staff members and then to remain at that level. Thewastage of recruits depends upon their length of service and is as follows:

End of week

Cost of individual replacement

Total cost of group replacement Average cost per week

1 100 × 2 = 200 1,000 × 0.50 +100×2 = Rs.700 Rs. 700.00 2 160× 2 = 320 700+ 160×2 = Rs. 1,020 Rs.510.00 3 281 ×2 = 562 1,020 + 281 ×2 = Rs. 1,582 Rs. 527.33

the end of year

(i) Find the number of staff members to be recruited every year.

(ii) If there are seven posts of Head of Deptt. for which length of service is the only creterion of promotion,what will be average length of service after which a new entrant should expect promotion?

Solution:

Let us assume that the recruitment per year is 100. From above it is clear that the 100 who join in thefirst year will become zero in 10th year, the 100 who join in the 2nd year will (serve for 9 years and)become 5 at the end of the 10th year and the 100 who join in the 3rd year will (serve or 8 years and)become 14 at the end of the 10th year and so on. Thus, when the equilibrium is attained, the distributionlength of service of staff members will be as under:

Year: 1 2 3 4 5 6 7 8 9 10 Total % who left up to: 5 35 36 65 70 76 80 86 95 100

Year No. of staff members 0 100 1 95 2 65 3 44 4 35 5 30 6 24 7 20 8 14 9 5

10 0 Total 432

∴


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(i) Thus if 100 staff members are recruited every year, the total number of staff members after 10 years ofservice = 432

100 To maintain a strength of 50, the number to be recruited every year = 432

100 x 50 = 11.6

It is assumed that those staff members who completed x years’ service but left before x + 1 years’service, actually left immediately before completing x + 1 years.

If it is assumed that they left immediately after completing x years’ service, the total number will

become 432 - 100 = 332 and 100 the required intake will be 50 x 332

100 = 15

In actual practice they may leave at any time in the year so that reasonable number of recruitments

per year = 2

156.11 + = 13 (approx)

(ii) If we recruit 13 persons every year then we want 7 seniors. Hence if we recruit 100 every year, we

shall require 13

7 x 100 = 54 (approx.) seniors.

It can be seen that 54 seniors will be available if we promote them during 6th year of their service ( ∵0 + 5 + 14 + 20 + 24 = 63 > 54 ).

∴ The promotion of a newly recruited staff member will be due after completing 5 years and beforeputting in 6 years of service.


Problem : 1. The following table gives the running costs per year and resale values of a certain equipmentwhose purchase price is Rs. 6,500. At what year is the replacement due optimally.

∴

Year 1 2 3 4 5 6 7 8 Running costs (Rs.) 1,400 1,500 1,700 2,000 2,400 2,800 3,300 3,900 Resale value (Rs.) 4,000 3,000 2,200 1,700 1,300 1,000 1,000 1,000

Year 1 2 3 4 5 6 Replacement cost at the beginning of the year 100 110 125 140 160 190 Salvage value at the end of the year 60 50 40 25 10 0 Operating costs 25 30 40 50 65 80

Problem: 2. A truck-owner finds from his past experience that the maintenance costs are Rs. 200 for thefirst year and then increase by Rs. 2,000 every year. The cost of the Truck Type A is Rs. 9,000. Determine thebest age at which to replace, i.e. truck. If the optimum replacement is followed what will be the averageyearly cost of owing and operating the Truck? Truck Type B cost Rs. 10,000. Annual operating costs are Rs.400 for the first year and then increase by Rs. 800 every year. The Truck owner have now the Truck Type Awhich is one year old. Should it be replaced with B type, and if so, when?

Problem:3. A Plant Manager is considering replacement policy to a new machine. He estimates the followingcosts:

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Find the year when replacement is to be made.

Problem : 4. A fleet owner finds from his past records that the costs per year of running a vehicle whosepurchase price is Rs 50,000 are as under:

Year 1 2 3 4 5 6 7Running cost Rs.) 5,000 6,000 7,000 9,000 11,500 16,000 18,000Resale value (Rs.) 30,000 15,000 7,500 3,750 2,000 2,000 2,000

Thereafter, running cost increases by Rs. 2,000, but resale value remains constant at Rs. 2,000. At what ageis a replacement due?

Problem : 5. The following mortality rates have been observed for a certain type of light bulbs:

There are 1,000 bulbs in sue, and it costs Rs. 2 to replace an individual bulb which has burnt out. If all bulbswere replaced simultaneously it would cost 50 paise per bulb. It is proposed to replace all bulbs at fixedintervals, whether or not they have burnt out, and to continue replacing burnt out bulbs as they fail. Atwhat intervals should all the bulbs be replaced?

Problem : 6. An electric company which generates and distributes electricity conducted a study on the lifeof poles. The repatriate life data are given in the following table:

Life data of electric poles

Week 1 2 3 4 5 Per cent failing by end of week 10 25 50 80 100

(i) If the Company now installs 5,000 poles and follows a policy of replacing poles only when they fail,how many poles are expected to be replaced each year during the next ten years?

To simplify the computation assume that failures occur and replacements are made only at the end ofa year.

(ii) If the cost of replacing individually is Rs. 160 per pole and if we have a common group replacementpolicy it costs Rs. 80 per pole, find out the optimal period for group replacement.

Problem : 7. A manufacturer is offered two machines A and B. A is priced at Rs. 5,000 and running costs areestimated Rs. 800 for each of the first five years, increasing by Rs. 200 per year in the sixth and subsequentyear. Machine B, which has the same capacity as A, costs Rs. 2,500 but will have running costs of Rs. 1,200per year for six years, increasing by Rs. 200 per year thereafter.

If money is worth 10% per year, which machine should be purchased? (Assume that the machines willeventually be sold for scrap at a negotiable price).

Years after installation:

1 2 3 4 5 6 7 8 9 10

Percentage poles failing:

1 2 3 5 7 12 20 30 16 4

∴


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Problem: 8. Suppose that a special purpose type of light bulb never lasts longer than 2 weeks. There is achance of 0.3 that a bulb will fail at the end of the first week. There are 100 new bulbs initially. The cost perbulb for individual replacement is Re. 1 and the cost per bulb for a group replacement is Re. 0.50. It ischeapest to replace all bulbs: (i) individually, (ii) every week, (iii) every second week, (iv) every thirdweek?

Problem: 9. A manufacturing firm is considering a policy of replacing certain key electrical componentsbelonging to one group of machines on a group replacement basis instead of making repairs as needed.There are approximately 100 parts of one type that have the mortality distribution shown below. The costof replacing parts on individual basis is estimated to be Rs. 10 per part whereas that on group basis comesto Rs. 3 per part. Compare the average weekly cost of the two replacement alternatives:

Week Probability of failure during week 1 0.3 2 0.1 3 0.1 4 0.2 5 0.3

Year Net capital cost (Rs.) (C–

S)

Running Cost (Rs.)

Cumulative running cost (Rs.)

Total cost Rs.

(2) + (4)

Average Annual cost (Rs.) (5)÷(1)

(1) (2) (3) (4) (5) (6) 1 2,500 1,400 1,400 3,900 3,900 2 3,500 1,500 2,900 6,400 3,200 3 4,300 1,700 4,600 8,900 2,967 4 4,800 2,000 6,600 11,400 2,850 5 5,200 2,400 9,000 14,200 2,840 6 5,500 2,800 11,800 17,300 2,882

Year

(1)

Capital cost (Rs.) (2)

Maintenance cost (Rs.)

(3)

Cumulative maintenance cost

(Rs.) (4)

Total cost (Rs.) (5) (2) + (4)

Avg. annual cost (Rs.) (6)

(5) ÷ (l)

1 9,000 200 200 9,200 9,200 2 9,000 2,200 2,400 11,400 5,700 3 9,000 4,200 6,600 15,600 5,200 4 9,000 6,200 12,800 21,800 5,450

Solution 1: Chart showing optimal Replacement period

∴ Optimal replacement period at the end of 5th year.

Solution 2 Truck A Chart Showing Optimal Replacement Period

∴

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Resource Management

Year Capital cost (Rs.)

Maintenance cost (Rs.)

Cumulative maintenance

cost (Rs.)

Total cost (Rs.)

Avg. annual cost (Rs.)

(1) (2) (3) (4) (2) + (4) (5) ÷ (1) 1 10,000 400 400 10,400 10,400 2 10,000 1,200 1,600 11,600 5,800 3 10,000 2,000 3,600 13,600 4,533 4 10,000 2,800 6,400 16,400 4,100 5 10,000 3,600 10,000 20,000 4,000 6 10,000 4,400 14,400 24,400 4,067

∴Optimal replacement period end of 3rd year.

Truck B Chart Showing Optimal Replacement Period

Comparing the average cost of Trucks A and B when the current truck A is one year old.

Average Annual Maintenance Cost

At the year third, the Truck A can be replaced by Truck B and subsequently it can be replaced at the end of5 years of its use.

Solution 3.

Chart Showing Optimal Replacement Period

Year A B 1 5,700 10,400 2 5,200 5,800 3 5,450 4,533

The asset should be replaced by the end of 2nd year where the average annual cost is lowest.

Year Net capital cost (Rs.) (Cost -

Scrap)

Operation cost (Rs.)

Cumulative operation Costs (Rs.)

Total cost (Rs.) Average annual cost (Rs.)

1 40 25 25 65 65 2 60 30 55 115 57.5* 3 85 40 95 180 60 4 115 50 145 260 65 5 150 65 210 360 72 6 190 80 290 480 80


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Solution 4.


Year (1)

Net capital cost (Rs.)

(2)

Annual maintenance cost

(Rs.) (3)

Cumulative operation costs

(Rs.) (4)

Total cost (Rs.)

(5)

Average annual cost

(Rs.) (6) (5) ÷ (l)

1 20,000 5,000 5,000 25,000 25,000 2 35,000 6,000 11,000 46,000 23,000 3 42,500 7,000 18,000 60,500 20,167 4 46,250 9,000 27,000 73,250 18,313 5 48,000 11,500 38,500 86,500 17,300 6 48,000 16,000 54,500 1,02,500 17,083* 7 48,000 18,000 72,500 1,20,500 17,214

Otimal replacement at the end of 6th year.

Solution 5.


It should be replaced by the end of Second Week.

Solution 6.


Average life of the pole 1 × 0.01 + 2 × 0.02 + 3 × 0.03 + 4 × 0.05 + 5 × 0.07 + 6 × 0.12 + 7 ×0.20 + 8 ×0.3 + 9 ×0.16 + 10 × 0.04 = 7.05

No. of poles to be replaced every year = 05.7

5000 = 709

Average yearly cost on individual replacement = 709×160 = Rs. 1,13,440.

Group Replacement: Initial Cost = 5,000 × 80 = 4,00,000

1 2 3 4 5 % failing during the week 10 15 25 30 20

Week Bulbs to be replaced cost Weekly cost

Cumulative cost

Total cost

Avg. Weekly

cost 1 1,000 × 0.1 = 100 200 200 700 700 2 1,000 ×0.15 + 100 × 0.1 = 160 320 520 1,020 510 3 1,000×0.25+100×0.15+160 ×0.10 =

281 562 1,082 1,582 527

Initial cost = 1000 x .5 = 500

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Resource Management

Optimal replacement at the end of the 5th year.

Solution 7.

Machine A: Chart Showing Optimal Replacement Period

Year No. of poles to be replaced Yearly cost

Cumulative cost

Total cost Avg.annual cost

1 5000 × 0.01 = 50 8,000 8,000 40,8000 40,8000 2 5000 × 0.02 +50 ×.01 = 101 16,160 24,160 42,4160 21,2080 3 5000 × 0.03+50 × 0.02

+101× 0.01= 152 24,320 48,480 44,8480 14,9493

4 5000 × 0.05 + 50 × 0.03+ 101× 0.02+152 × 0.01 = 256

40,960 89,440 48,9440 12,2360

5 5,000 × 0.07 + 50 × 0.05 + 101 × 0.03 + 152 × 0.02 + 265 × 0.01 = 362

57,920 1,47,360 5,47,360 1,09,472

6 5,000 × 1.2 + 50 × 0.07 + 101 × 0.05 + 152 × 0.03 + 256 × 0.02 + 362 × 0.01 = 6023

9,63,680 11,11,040 15,11,040 2,51,840

Year Running cost (Rs.)

Discount factor

Discounted running

Cost (Rs.)

Cumulative running Cost

(Rs.)

Total cost (Rs.)

Cumulative discounted

factor

Annualized cost (Rs.)

(1) (2) (3) (4) (5) (6) (7) (8) = (6) ÷7 1 800 1.00 800 800 5,800 1.00 5,800 2 800 0.9091 727 1,527 6,527 1.9091 3,419 3 800 0.8264 661 2,188 7,188 2.7355 2,628 4 800 0.7513 601 2,789 7,789 3.4868 2,234 5 800 0.6830 546 3,335 8,335 4.1698 1,999 6 1,000 0.6209 621 3,956 8,956 4.7907 1,869 7 1,200 0.5645 677 4,633 9,633 5.3552 1,799 8 1,400 0.5132 718 5,351 10,351 5.8684 1,764 9 1,600 0.4665 746 6,097 11,097 6.3349 1,752 10 1,800 0.4241 763 6,860 11,860 6.7590 1,755


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Year Running cost (Rs.)

Discount factor

Discounted running

Cost (Rs.)

Cumulative running

Cost (Rs.)

Total cost (Rs.)

Cumulative discounted

factor

Annualized cost (Rs.)

(1) (2) (3) (4) (5) (6) (7) (8)=6÷7 1 1,200 1.00 1,200 1,200 3,700 1.0000 3,700 2 1,200 0.9091 1,091 2,291 4,791 1.9091 2,510 3 1,200 0.8264 992 3,283 5,783 2.7353 2,114 4 1,200 0.7513 902 4,184 6,684 3.4868 1,917 5 1,200 0.6830 820 5,004 7,504 4.1698 1,800 6 1,200 0.6209 745 5,750 8,250 4.7907 1,722 7 1,400 0.5645 790 6,540 9,040 5.3552 1,688 8 1,600 0.5132 821 7,361 9,861 5.8684 1,680 9 1,800 0,4665 840 8,201 10,701 6.3349 1,689

Machine B:

Chart showing optimal Replacement period

Machine B should be purchased as its annual cost is lowest i.e. Rs. 1,680.

Solution 8.


Probability 0.3; 0.7

Individual replacement: Average life of bulb = 1 × 0.3 + 2 × 0.7 = 1.7

Weekly replacement = 1001.7

= 59

Group replacement: 100 × 0.5 = 50

Week Items to be replaced Weekly cost

Cumulative cost

total cost Avg. weekly cost

1 100 × 0.3 = 30 30 30 80 80 2 100 × 0.7 + 30 × 0.3 = 79 79 109 159 79.50

It should be replaced by the end of second week. It cannot be replaced every third week as the life of thebulb is only 2 weeks.

Solution: 9 Replacement cost on individual basis 5

000,1.10100 Rs=× = Rs. 200 per week. Replacement

cost on group basis 3(100 × 0.3) + 3(100 × 0.1) + 3(100 × 0.1) + 3(100 × 0.2) + 3(100 × 0.3) = 300/5 = Rs.60/-

4.4 Change of Technology and its Implication

Technology is a resource of profound importance not only in production but also to profitability andgrowth of the entire business organisation. Technology drives productivity and also drives change in thisworld. Technological change is a major factor in gaining competitive advantage. The development andInnovative use of technology can provide a firm a distinctive competence. Competitive advantage can beachieved not just from creating new technology but also by applying and integrating existing technologies.

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Resource Management

Technology is a significant ingredient in virtually all production and operations management decisions.Advances in computer technologies (both hardware and software) automation, robotics, lasers, andinformation and communication technologies have had broad-reaching impact across all industries. Tostay competitive, manufacturing and service organisations must adopt new technologies.

Firms that have used technology as a competitive weapon have effectively integrated their technologystrategy and business strategy. As these firms invent and develop new technologies, they offer new productsand services.

Technology is defined to be know-how, physical things and procedures used to produce products andservices. Know-how means knowledge and judgement of how, when and why to employ equipments,processes and procedures. Knowledge includes craftsmanship and experience, physical things areequipments and tools, procedures are the rules and techniques for operating the equipment and performingthe work.

Technologies require a support network to be implemented. A support network consists of physical,informational and organisational relationships that make a technology complete and allow it to functionas intended.

Advanced technology refers to the application of the latest scientific or engineering discoveries to thedesign of production and operations processes.

Technology and technique do not mean the same. Technique is the totality of methods rationally arrived atand having absolute efficiency whereas technology is the organisation and application of knowledge forthe achievement of practical purposes. Some of the examples of technology are manufacturing or productiontechnology, design technology, computer technology, communication technology, nuclear technology,satellite; communication technology, space technology, missile technology, laser technology and the like.

Advances in technologies create new products and services and reshape processes. Technology takes manyforms, beginning with ideas, knowledge, and experience and then utilising them to create new sad betterways of doing things. Technology provides distinctive competency and competitive advantage to a firmover others. The impact of technology is pervasive.

A vital factor in production organisations is whether the technology involves capital intensive operation(i.e., large investment in plant and machinery) or labour-intensive operations. Technology will affect: (i)the organisation of production, (ii) the capital investment in plant and equipments buildings etc., (iii) thescale (or volume) of production operations, (iv) the influence of labour relations in-production operations.

The management of production is vitally concerned with the technology of the production processes. Itmust organise according to the technology adopted and by the adoption of productive system which iseither capital intensive or labour intensive. Also, it must design a highly sophisticated production controlsystem for batch production, or a material management system for high capacity assembly line operations.

The organisation is not simply a technical or social system. It requires structuring and integrating humanactivities around various technologies. The technical system is determined by the task requirements andshaped by the specialisation of knowledge and skills required, the types of machinery and equipmentsinvolved, the information processing requirements and the layout of facilities.

Any change in the technical system affects the organisational elements. The impact of technology on theorganisation-its goals, structure, psychological system and managerial system will be quite significant.


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Technology can be classified as: (i) manual technology, (ii) mechanised technology (or mechanisation), (iii)Automated technology (or automation), (iv) current technology, (v) appropriate technology, (vi) state-of-the-art technology, (vii)advanced technology, (viii) obsolete technology (or out-dated technology), (ix)capital-intensive technology, (x) labour intensive technology, etc.

Manual technology is the use of muscular power to do work which was prevalent before industrialrevolution. Mechanised technology or mechanisation uses machine power in place of muscle power and isthe first step towards automation. Examples of mechnisation are power operated tools, tool changingdevices, powered materials handling equipments such as conveyors, jib cranes for loading and unloadingheavy jobs etc. Automated technology or automation is any form of equipment or machine which willcarryout a preset program or sequence of operations and at the same time measure and correct its actualperformance in relation to that program. Current technology is any technology currently used by a firm inits operations. Appropriate technology is the technology which meets the requirements of time, place andobjectives of a firm at a particular point of time. Appropriateness is an inherent quality in technology.Depending on the social, political, economic and other conditions prevailing in a country at a particularpoint of time and also depending on the priorities of the country, a technology becomes appropriate eventhough it may not be an advanced technology or state-or-the-art technology. (For example, whether capital-intensive or labour-intensive technology is the appropriate technology to be adopted by a firm or industryor a country).

State-of-the-art technology is a modern technology which has been adopted by many developed countriesin the world. This technology will enable the firm to produce state-of the art products using state-of-the artdesign. It is also known as proven technology. Advanced technology is the latest technology based on thelatest scientific or engineering discoveries and used in the design and production processes. It is alsoknown as new technology which is developed by a firm which is a technology leader in its field of operation.Examples of advanced technologies are space technology, missile technology, information technology,laser technology, bio-technology and so on.

Obsolete technology is an out-dated technology which has been replaced by a superior technology, therebyresulting in obsolescence of the old technology which has been replaced.

Capital-intensive technology involves huge investments in capital assets such as equipment and machinery,materials handling and storage systems, information handling (storage and retrieval) systems,communication systems and office automation equipments.

Labour-intensive technology does not involve investment in huge capital intensive systems but make useof abundant labour (manpower) available in the country. For example, in India, textile and mining Industriesadopt labour-intensive technologies.

The Choice of Technology: The choice of technology depends on several factors, both internal and externalto the organisation choosing the technology. The various internal factors are : (i) availability of funds forinvestment, (ii) product life cycle and technology-life cycle position, (iii) present plant capacity andtechnology adopted (i.e., current technology).

Technology can be quite capital intensive and require high investment in equipments, machines andprocesses. The question is whether the firm can afford to invest in a new technology which may be highlyexpensive. Also, the change to a new technology is not advisable when the product is in the saturation ordecline stage in its life cycle. New technologies for processing may be best suited for developing andmanufacturing new products. The new technology chosen should be capable of matching with the existing

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technology and plant capacity so that there will be a synergy effect on the plant capacity when a newtechnology is adopted.

The external factors involved in the choice of technology are: (i) Government polices and regulations,availability of resources such as raw materials, energy, skilled labour etc., required for using the newtechnology, (ii) market scenario (market demand, customer requirement of product quality etc.).

Technology Life-Cycle: Like a product has its life cycle, technology also has a life cycle. The various phasesor stages in a technology life cycle are:

(i) Innovation in which stage a new technology (product or process technology) is developed, (ii)Syndication during which stage, the technology is demonstrated and slowly commercialised, (iii)Diffusion stage in which a new technology gradually replaces the current technology and (iv)Substitution stage in which the current technology becomes obsolete and is completely replaced bythe new technology.

The recent development in the technology of product design comprises certain tools provided by theinformation sciences that contribute to better, cheaper and faster design of products. They are:

(i) Computer-aided design (CAD)

(ii) Computer-aided design and manufacture (CAD/CAM).

(i) Computer Aided Design (CAD): It is an electronic system using computers for designing new partsor products or modifying existing ones, replacing the traditional drafting work done by a draftsmanon a drafting board. The CAD consists of a powerful desktop computer and graphics software thatenables a designer to manipulate geometric shapes. The designer can create drawings and view themfrom any angle on a display monitor. CAD softwares have been developed for designing electroniccircuits, printed-circuit-board design, designing and drafting three dimensional drawings and alsofor analysis of heat and stress in mechanical designs.

Advantages of CAD are:

(i) Allows designers to save time and money by shortening design and development cycle time, (ii) Eliminatesprototype model building to prove the design, (iii) Allows designers to determine costs and test suchvariable as stress, tolerance, product variability, interchangeability and serviceability, (iv) Low cost ofdesign even for a custom-built, low volume product,(v) Eliminates manual drafting completely,(vi) Makesreview of numerous options in design possible before final commitments are made because of the speedand ease with which sophisticated designs can be manipulated, (vii) Faster development, better productsand accurate flow of information to other departments and (viii) Product cost can be determined at thedesign stage itself.

Production technology

In addition to the developments in design technology, a number of advances are made in technology usedto enhance production. Some these advancements in production technology are:

(i) Numerical control and Computer Numerical Control (NC and CNC), (ii) Automated process controls,(in) Vision systems (automated inspection systems), (iv) Robots, (v) Automated identification systems(AIS), (vi) Automated storage and retrieval system (ASRS), (vii) Automated guided vehicles (AGV),(viii) Automated Sow lines, (ix) Automated assembly systems, (x) Flexible manufacturing systems(FMS), (xi) Computer integrated manufacturing (CIM), (xii) Enterprise resource planning.


283

The above aspects of production technologies are discussed in detail in the following section.

(i) Numerical Control and Computer Numerical Control: Many machines such as lathe, milling, drillingand boring machines are now designed for electronic control called numerical control (NC). Thenumerically controlled (NC) machines have control systems which read instructions and translatethem into machine operations. When these machines are programmed through their ownminicomputers and have memories to store these programs, they are called computer numerical control(CNC) machines.

Advantages of N/C Machines: (i) Smaller machine setup time, (ii) Machine motions and tool changingare controlled by instructions on a control system, (iii) Increased productivity and higher quality, (iv)Suitable for low volume production.

Advantages of CNC Machines: (i) Instructions may be stored and handled more efficiently, ii) Microcomputer system controls the machine settings and operations rather than human beings, (iii) Realtime and off-line diagnostic possibilities may be built into the CNC system, (iv) Machining data andoperator instructions may be displayed on the screen of computer.

Application of NC/CNC Machines : These machines are used to machine parts (i) with complexmachining requirements, (ii) requiring high precision, (iii) in developmental stages where many changesmay be needed, (iv) normally requiring extensive tooling, (v) requiring fast or slow machining speeds,(vi) made from expensive raw materials and (vii) required in small quantities of repetitive batches.

(ii) Automated Process Control: Automated process control makes use of information technology tomonitor and control a physical process. It is also used to determine and control temperatures, pressuresand quantities in petroleum refineries, cement plants, chemical plants, steel mills, nuclear reactorsetc., process control systems operate in a number of ways : (a) Sensors collect data, (b) Analog devicesread data periodically (say once a minute or once a second), (c) Measurements are translated intodigital signals which are transmitted to a digital computer, (d) Computer programs read the digitaldata and analyse the same, (e) The resulting output may take numerical forms which include messageson computer consoles or printers, signal to motors to change valve settings, warning lights or sirensetc.

(iii) Vision Systems (Automated Inspection Systems): These are machines or equipments that have beenintegrated into the inspection of products for controlling quality. They combine video camera andcomputing technology to take physical dimensions of parts, compare the measurements to standardsand determine whether the parts meet quality specifications. Vision systems are also used for visualinspection in food processing organisations. Automated inspection systems facilitate 100 percentinspection which will lead to improved product quality and reduced inspection costs.

(iv) Robots: Robotics or robotry is a fast developing field of technology in which human like machinesperform production tasks. A robot is a reprogrammable, multifunctional manipulator designed tomove materials, parts, tools or specialised devices through variable programmed motions for theperformance of a variety of tasks. Robots are machines which are flexible, have the ability to hold,more and grab items. They are controlled by micro computers which when programmed guide themachines through their predetermined operations.

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Advantages:

(i) do not strike work, (ii) do not mind hot, dirty, dusty working conditions, (iii) can work at highspeeds, (iv) will not sue if injured, (v) can work long hours without breaks, (vi) can be used for welding,painting, assembly work, loading, unloading, material handling and other repetitive, monotonouswork.

(v) Automated Identification Systems (AIS): These use bar codes, radio frequencies, magnetic stripes,optical character recognition and machine vision to sense and input data into computers. These systemsreplace human beings to read data from products, documents, parts and containers and interpret thedata. An example is the system used to identify and read the bar code on an item in the check-outcounters at grocery stores. A scanner reads the identification number from the bar code on the itemaccesses a computer data base, and sends the price of the item to the cash register and updates the ;inventory data in the inventory system.

(vi) Automated Storage and Retrieval Systems (ASRS): Computer controlled warehouses use ASRS;which provide for the automatic placement and withdrawal of parts and products into and fromdesignated storage places in the warehouse. Such systems are commonly used in distribution facilitiesof retailers.

(vii) Automated Guided Vehicles (AGVs): These are automated materials handling and delivery systemswhich can take the form of mono-rails, conveyors, driverless trains, pallet trucks and unit load carriers.AGVs are electronically guided and controlled vehicles used to move parts and equipments. AGVsusually follow either embedded guide wires or paint strips through operations until their destinationsare reached.

(viii) Automated Flow Lines: An automated flow line includes several automated machines which arelinked by automated transfer machines and handling machines. The raw material feeders automaticallyfeed the individual machines and operations are carried out without human intervention. After anitem is machined on one machine on the line, the partially completed item is automatically transferredto the next machine on the line in a predetermined sequence, until the job is completed. Majorcomponents such as automobile rear axle housings are produced using automated flow lines. Thesesystems are also known as fixed automation or hard automation because the flow lines are designedto produce only one type of component or product. These systems are suitable for products with highand stable demand because of very high initial investment required and the difficulty of changingover to other products. But production systems which provide greater flexibility (For example, FlexibleManufacturing Systems) are more favoured than fixed automation.

(ix) Automated Assembly Systems: In this system automated assembly machines or equipments are linkedtogether by automated materials handling equipments. Examples of automated assembly equipmentsare robotic welders or component insertion unit which are used to join one or more parts or componentsor assemblies. The partially assembled product is moved to the next assembly equipment automaticallyby the automated materials handling equipments and this process is repeated until the whole assemblyis completed. Automated assembly systems require unique product design which is suitable to thesesystems unlike product designs suitable for manual assembly operations. Some of the principles thatare applied while designing products for automated assembly are:

(i) The amount of assembly required must be reduced: For example, use of castings or plasticmouldings (single part) instead of making the part by assembling two or more sheet metal partsfastened together by bolts and nuts or by rivets.


285

(ii) The number of fasteners required must be reduced: Joining two or more parts by welding ispreferred to joining by fasteners such as screws, rivets, bolts and nuts. Product designs shouldfacilitate this.

(iii) Components must be designed to be automatically delivered and positioned: Hoppers, slottedchutes, vibratory bowls etc. are used to feed the part and orient them to deliver to the assemblyequipment and position in the proper place for assembly to be done automatically. Productdesigns should facilitate this.

(iv) Products must be designed for layered assembly and vertical insertion of parts: Product designmust facilitate build up of assembly from a base upward in layers and the components insertedvertically into the assembly.

(v) Parts must be designed such that they are self aligning: Design should facilitate aligning of partsas and when they are inserted into assemblies.

(vi) Products must be designed into major modules for production (modular design): Each moduleis assembled in an automated assembly system and them the modules are assembled into thefinal product.

(vii) Quality of components should be high: High quality components avoid jams in feeding andassembly mechanisms. The advantages of automated assembly units are low production costper unit, high product quality and higher product flexibility.

(x) Flexible Manufacturing System (FMS): A flexible manufacturing system (FMS) is a configuration ofa group of production machines (or workstations) connected by automated material handling andtransferring machines and integrated by computer system which can give instructions to producehundreds of different parts in whatever order specified.

An FMS is a type of flexible automation system which builds on the programmable automation of NCand CNC machines. Materials are automatically handled and loaded and unloaded for machiningoperations. Programs and tooling set up can be quickly changed and production can be quicklyswitched on from one job to another with no loss of change overtime.

Key components of an FMS are:

(i) Several computers controlled machining centres or workstations having CNC machines and robotsfor loading and unloading.

(ii) Computer controlled transport system (AGVs) for moving materials and parts from one machine toanother and in and out of the system.

(iii) Computer controlled robots for loading and unloading stations.

(iv) An automated storing and retrieval system.

All the above subsystems of FMS are controlled by a central computer with the needed software. Rawmaterials are loaded on the AGVs which bring them to the work centres as per the sequence ofoperations unique to each part. The route is determined by the central computer. The robots lift thematerials from the AGV and place on the workstation, where the required operations are carried out.After the completion of operations, the robots unload the job and place it on the AGV to move the jobto the next workstation as per the sequence of operations.

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The FMS is suitable for intermediate flow strategies with medium level of product varieties and volumes(40 to 2000 units per part). Also, FMS can produce low variety high volume products in the same wayas fixed automation systems.

Advantages: (i) Improved capital utilisation, (ii) Lower direct labour cost, (iii) Reduced inventory,(iv) Consistent quality, (v) Improved productivity.

Disadvantage: (i) High initial capital investment, (ii) Limited ability to adopt to product changes, (iii)Substantial preplanning, tooling and fixture requirements, (iv) Standardisation of part designs neededto reduce number of tools required, (v) Requires long planning and development cycle to install FMS.

(xi) Computer Integrated Manufacturing (CIM): Computer Integrated Manufacturing is a system whichacts as a bridge or umbrella to integrate product design and engineering, process planning andmanufacturing using complex computer systems. It integrates CAD, CAM, FMS, inventory control,warehousing and shipping. A computer-aided drafting generates the necessary electronic instructionsto run a direct numerically controlled (DNC) machine and any design change initiated at a CADterminal can incorporate that change in the part produced on the shop floor within a few minutes.When this kind of capability is integrated with inventory control, warehousing and shipping as a partof a flexible manufacturing system, the entire system is called computer-integrated-manufacturing.

The CIM systems work as follows:

1. Top Management decides to make a product based on market opportunities, the company’s strengthand weaknesses and formulates its strategic plan based on competitive advantage.

2. Production/Operations Management implements production process, supplier coordination, materialplanning, scheduling operations, delivery schedules and cost controlling functions.

3. Computer-Aided-Design facilitates the design of the product, its quality analysis, planning themanufacturing process, designing tools and. fixtures and machine loading programs.

4. Computer-Aided-Manufacturing allows fabrication of raw materials into parts to be sent to theassembly lines.

5. Automated Storage and Retrieval System (ASRS) and automated guided vehicles (AGVs) facilitatestorage/retrieval, movement of incoming materials and parts, work-in-process and finished products

Computer Integrated Manufacturing System

Top Management

Management Information

System (MIS)

Productional Operations

Management (PQM)

Computer Aided

Manufacturing (CAM)

Computer Aided Design (CAD)

Robots Automated Storage

and Retrieval System (ASRS) and

Automated Guided vehicles (AGVS)


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6. Robots assemble the subunits into final products, test them with automated equipments and put thefinished products into containers for shipment.

7. Management Information System (MIS) integrates all the elements of the computer integrated systemsamong themselves and also with the top management of the organisation.

(xii) Enterprise Resource Planning (ERP) System: ERP systems comprise latest comprehensive softwarepackages to automate a number of business processes. This software integrates most of the businessfunctions in an organisation. ERP systems have automated manufacturing processes, organised accountbooks, streamlined corporate departments such as human resources and facilitate business processre-engineering.

ERP systems need complex set of software programs and heavy investment to implement them. SeveralERP software packages have been developed by leading software companies such as SAP, Oracle,J.D. Edwards, People Soft and Baan. The latest development in ERP system has been the integrationof e-business capabilities.

E-business uses the internet to conduct or facilitate business transactions such as sales, purchasing,communication, inventory management, customer service, placing purchase orders and checking thestatus of purchase orders etc. ERP software packages were modified with additional features to facilitatee-business.

Automation Issues

Some of the issues to be deliberated in the use of automation are:

(i) Not all automation projects are successful.

(ii) Automation is not a substitute for poor management.

(iii) Some automation projects may not be worth-while based on economic analysis.

(iv) Some operations are not technically feasible to be automated.

(v) Small and start-up business may not be able to invest for automation projects.

Advantages and Disadvantages of Automation

Advantages: (i) Increased output and higher productivity, (ii) Improved and uniform quality, (iii) Reducedcosts, (iv) Fewer accidents, (v) Better production control, (vi) Dangerous and unpleasant taste can beperformed by robots.

Disadvantages: (i) Heavy capital investment, (ii) Displacement of labour, (iii) Loss of suggestions fromemployees, (iv) Design specifications for raw materials can not be relaxed, (v) Cost of shutdown of automatedplant due to shortage of materials is quite high, (vi) Dehumanisation.

Factories of the Future

The features of factories of the future are:

1. Stress on high product quality.

2. Greater emphasis on flexibility.

3. Faster execution and delivery of customer orders.

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Resource Management

4. Change in production economics - fixed costs will become variable and variable costs will change tofixed costs. Product will be designed using CAD/CAM which forms the basis for process planningalso. CIM systems will be extensively used.

5. Organisational structure changes with line personnel becoming staff personnel and vice versa. Themainstream activities will be maintenance, quality assurance, product design and engineeringmanaging technology change, software development etc.

6. The factories of future will be driven by computers used in CIM systems.

Objective TypeQuestions and

Answer

290

Objective Type Questions and Answers

Type: - Choosing of Correct Answers:

1 Application of technology or process to the raw material to add use value is known as:

(a) Product, (b) Production, (c) Application of technology, (d) Combination of technology and process.

Ans. (b)

2. Surface hardening is an example of:

(a) Production by application of machine tool, (b) Production by disintegration, (c) Production byIntegration, (d) Production by Service.

Ans.(d)

3. In Production by disintegration the material undergoes:

(a) Change in economic value only, (b) Change in physical and chemical characteristics, (c) Change intechnology only, (d) None of the above.

Ans.(b)

4. In Production by service, the product undergoes the changes in:

(a) Shape and size of the surface, (b) Shape of the surface only, (c) Size of the surface only,

(d) Chemical and Mechanical properties.

Ans.(d)

5. Use of any process or procedure designed to transform a set of input elements into a set of outputelements is known as:

(a) Transformation process, (b) Transformation of input to output, (c) Production, (d) Technologychange.

Ans.(c)

6. Conversion of inputs into outputs is known as:

(a) Application of technology, (b) Operations management, (c) Manufacturing products, (d) Product.

Ans.(b)

7. The desired objective of Production and Operations Management is:

(a) Use cheap machinery to produce, (b) To train unskilled workers to manufacture goodsperfectly, (c) Optimal utilisation of available resources, (d) To earn good profits.

Ans.(c)

8. The scope of Production Planning and Control is:

(a) Limited to Production of products only, (b) Limited to production of services only, (c) Limited toproduction of services and products only, (d) Unlimited, can be applied to any type of activity.

Ans.(d)


291

9. Manufacturing system often produces:

(a) Standardised products, (b) Standardised products in large volumes, (c) Substandardproducts inlarge volumes, (d) Products and services in limited volume.

Ans.(a)

10. The difference between product system and project system is:

(a) Project system the equipment and machinery are fixed where as in product system they are movable,(b) In Product system the machinery and equipment are fixed and in project system they are not fixed,(c) Project system produces only standardized products and product system produces onlyunstandardised products, (d) Products cannot be stocked whereas projects can be stocked.

Ans.(b)

11. Most important benefit to the consumer from efficient production system is

(a) He can save money, (b) He will have product of his choice easily available, (c) He gets increaseduse value in the product, (d) He can get the product on credit.

Ans.(c)

12. Two important functions that are to be done by Production department are:

(a) Forecasting, (b) Costing, (c) Scheduling and loading, (d) Inspecting.

Ans.(c)

13. Fixing the flow lines of materials in production is known as:

(a) Scheduling, (b) Loading, (c) Planning, (d) Routing.

Ans.(d)

14. The act of releasing the production documents to the production department is known as:

(a) Planning, (b) Routing, (c) Dispatching, (d) Releasing.

Ans.(c)

15. The activity of specifying when to start the job and when to end the job is known as:

(a) Plaining, (b) Scheduling, (c) Timing, (d) Follow-up.

Ans.(b)

16. In an organization, generally the production management is a

(a) Line function, (b) Staff function, (c) Line and Staff function, (d) None of the above.

Ans.(a)

17. In an organisation the production planning and control department comes under:

(a) Planning department, (b) Manufacturing department, (c) Personal department,

(d) R & D department.

Ans.(b)

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18. In Job production system, we need:

(a) More unskilled labours, (b) Skilled labours, (c) Semi-skilled labours, (d) Old people.

Ans.(b)

19. In Continuous manufacturing system, we need:

(a) General purpose machines and Skilled labours, (b) Special machine tools and highly skilledlabours, (c) Semi automatic machines and unskilled labours, (d) General purpose machines andunskilled labours.

Ans.(b)

20. Most suitable layout for Job production is:

(a) Line layout, (b) Matrix layout, (c) Process layout, (d) Product layout.

Ans.(c)

21. Most suitable layout for Continuous production is:

(a) Line layout, (b) Process Layout, (c) Group technology, (d) Matrix layout.

Ans.(a)

22. One of the product examples for Line layout is:

(a) Repair workshop, (b) Welding shop, (c) Engineering College, (d) Cement.

Ans.(d)

23. Number of product varieties that can be manufactured in Job production is:

(a) Limited to one or two, (b) Large varieties of products, (c) One only, (d) None of the above.

Ans.(b)

24. Number of product varieties that can be manufactured in Mass production is:

(a) One only, (b) Two only, (c) Few varities in large volumes, (d) Large varities in small volumes.

Ans.(c)

25. In general number of product varities that can be manufactured in Flow production is:

(a) One only, (b) Ten to twenty varities, (c) Large varities, (d) Five only.

Ans.(a)

26. Generally the size of the order for production in Job production is

(a) Small, (b) Large, (c) Medium, (d) Very large.

Ans.(a)

27. Generally in continuous production the production is carried out to:

(a) Customer’s order, (b) Government orders only, (c) For stock and supply, (d) Few rich customers.

Ans.(c)


293

28. Inventory cost per product in intermittent production is:

(a) Higher, (b) Lowest, (c) Medium, (c) Abnormal.

Ans.(a)

29. The material handling cost per unit of product in Continuous production is:

(a) Highest compared to other systems, (b) Lower than other systems, (c) Negligible, (d) Cannot say.

Ans.(b)

30. Routing and Scheduling becomes relatively complicated in

(a) Job production, (b) Batch production, (c) Flow production, (d) Mass production.

Ans.(b)

31. The starting point of Production cycle is:

(a) Product design, (b) Production Planning, (c) Routing, (d) Market research.

Ans.(d)

32. Variety reduction is generally known as:

(a) Less varities, (b) Simplification, (c) Reduced varities, (d) None of the above.

Ans.(b)

33. Preferred numbers are used to:

(a) To determine the number of varities that are to be manufactured, (b) To the test the design of theproduct, (c) To ascertain the quality level of the product, (d) To evaluate the production cost.

Ans.(a)

34. The act of assessing the future and make provisions for it is known as

(a) Planning, (b) Forecasting, (c) Assessment, (d) Scheduling.

Ans.(b)

35. For a marketing manager, the sales forecast is:

(a) Estimate of the amount of unit sales or a specified future period, (b) Arranging the sales men todifferent segments of the market, (c) To distribute the goods through transport to satisfy the marketdemand, (d) To plan the sales methods.

Ans.(a)

36. The time horizon selected for forecasting depends on:

(a) The salability of the product, (b) The selling capacity of Salesman, (c) Purpose for which forecast ismade, (d) Time required for production cycle.

Ans. (c)

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37. For production planning:

(a) Shot term forecasting is useful, (b) Medium term forecasting is useful, (c) Long term forecasting isuseful, (d) Forecasting is not useful.

Ans.(a)

38.. In general, medium range forecasting period will be approximately:

(a) 5 to 10 Years, (b) 2 to 3 days, (c) 3 to 6 months, (d) 10 to 20 years.

Ans.(c)

39. The range of Long range forecasting period may be approximately:

(a) 1 to 2 weeks, (b) 2 to 3 months, (c) 1 year, (d) above 5 years.

Ans.(d)

40. To plan for future man power requirement:

(a) Short term forecasting is used, (b) Long range forecasting is used, (c) Medium range forecasting isused, (d) There is no need to use forecasting, as future is uncertain.

Ans.(b)

41. Long range forecasting is useful in:

(a) Plan for Research and Development, (b) To Schedule jobs in Job production, (c) In purchasing thematerial to meet the present production demand, (d) To assess manpower required in the comingmonth.

Ans.(a)

42. Medium range forecasting is useful in:

(a) To assess the loading capacity of the machine, (b) To purchase a materials for next month, (c) Toplan for-capacity adjustments, (d) To decide whether to receive production orders or not.

Ans.(c)

43. To decide work load for men and machines:

(a) Medium range forecasting is used, (b) Short term forecasting is used, (c) Long range forecasting isused, (d) A combination of long range and medium range forecasting is used.

Ans.(b)

44. Important factor in forecasting production is:

(a) Environmental changes, (b) Available capacity of machines, (c) Disposable income of the consumer,(d) Changes in the preference of the consumer.

Ans.(b)

45. Technological development is an:

(a) Upward trend, (b) Downward trend, (c) Seasonal trend, (d) Erratic trend.

Ans.(a)


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46. A method in which a trend line drawn in such a way that the sum of the squares of deviations of theactual points above and below the trend line is at the minimum is known as:

(a) Squared trend method, (b) Equal square method, (c) Adjusted square method, (d) Least Squaremethod.

Ans.(d)

47. One of the advantages of Method of Least square is:

(a) It is a very easy method, (b) It does not use mathematics, (c) Trend values of all years of the seriesmay be obtained, (d) None of the above.

Ans.(c)

48. Line of Best fit is another name given to:

(a) Method of Least Squares, (b) Moving average method, (c) Semi average method, (d) Trend linemethod.

Ans.(a)

49. One of the important basic objectives of Inventory management is:

(a) To calculate EOQ for all materials in the organisation, (b) To go in person to the market andpurchase the materials, (c) To employ the available capital efficiently so as to yield maximum results,(d) Once materials are issued to the departments, personally check how they are used.

Ans.(c)

50. The best way of improving the productivity of capital is:

(a) Purchase automatic machines, (b) Effective Labour control, (c) To use good financial management,(d) Productivity of capital is to be increased through effective materials management.

Ans.(d)

51. MRP stands for:

(a) Material Requirement Planning, (b) Material Reordering Planning, (c) Material RequisitionProcedure, (d) Material Recording Procedure.

Ans.(a)

52. JIT stands for:

(a) Just in time purchase, (b) Just in time production, (c) Just in time use of materials, (d) Just in timeorder the material.

Ans.(b)

53. The cycle time, selected in balancing a line must be:

(a) Must be greater than the smallest time element given in the problem, (b) Must be less than thehighest time element given in the problem, (c) Must be slightly greater than the highest time elementgiven in the problem, (d) Left to the choice of the problem solver.

Ans.(c)

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54. The lead-time is the time:

(a) To placeholders for materials, (b) Time of receiving materials, (c) Time between receipt of materialand using materials, (d) Time between placing the order and receiving the materials.

Ans.(d)

55. Production planning deals with:

(a) What production facilities is required and how these facilities should be laid out in space available,(b) What to produce and when to produce and where to sell, (c) What should be the demand for theproduct in future? (d) What is the life of the product?

Ans.(a)

56. The first stage in production planning is:

(a) Process Planning, (b) Factory Planning, (c) Operation Planning, (d) Layout planning.

Ans.(b)

57. In Process Planning we plan:

(a) Different machines required, (b) Different operations required, (c) We plan the flow of material ineach department, (d) We design the product.

Ans.(c)

58. In Operation Planning:

(a) The planner plans each operation to be done at work centers and the sequence of operations,(b) Decide the tools to be used to perform the operations, (c) Decide the machine to be used to performthe operation, (d) Decide the materials to be used to produce the product.

Ans.(a)

59. Before thinking of routing, the production planner has to:

(a) Decide the optimal allocation of available resources, (b) To decide what type of labour to be used,(c) To decide how much of material is required, (d) To count how many orders he has on his hand.

Ans.(a)

60. The quantities for which the planner has to prepare production plan are known as:

(a) Optimal quantity of products, (b) Material planning, (c) Quantity planning, (d) Planning quantitystandards.

Ans.(d)

61. The document, which is used to show planning quantity standards and production plan, is known as:

(a) Planning specifications, (b) Route sheet, (c) Bill of materials, (d) Operation sheet.

Ans.(a)


297

62. One of the seven routing decisions is:

(a) Where to purchase the material?, (b) How to purchase material?, (c) When to purchase the material?(d) To make the product or buy the product?

Ans.(d)

63. In route sheet or operation layout, one has to show:

(a) A list of Materials to be used, (b) A list of machine tools to be used, (c) Every work center and theoperation to be done at that work center, (d) The cost of product.

Ans.(c)

64. The cycle time in selected in balancing a line must be:

(a) Must be greater than the smallest time element given in the problem, (b) Must be less than thehighest time element given in the problem, (c) Must be slightly greater than the highest time elementgiven in the problem, (d) Left to the choice of the problem solver.

Ans.(c)

65. In solving a problem on LOB, the number of workstations required is given by:

(a) Cycle time/Total time, (b) Cycle time/Element time, (c) Total time/Element time, (d) Total time/Cycle time.

Ans.(d)

66. (Total station time/Cycle time × Number of work stations) × 100 is know as:

(a) Line Efficiency, (b) Line smoothness, (c) Balance delay of line, (d) Station efficiency.

Ans.(a)

67. Final stage of production planning, where production activities are coordinated and projected on atime scale is known as:

(a) Scheduling, (b) Loading, (c) Expediting, (d) Routing.

Ans.(a)

68. Scheduling shows:

(a) Total cost of production, (b) Total material cost, (c) Which resource should do which job and when,(d)The flow line of materials.

Ans.(c)

69. Scheduling deals with:

(a) Number of jobs to be done on a machine, (b) Number of machine tools used to do a job, (c) Differentmaterials used in the product, (d) Fixing up starting and finishing times of each operation in doing ajob.

Ans.(d)

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70. The study of relationship between the load on hand and capacity of the work centers is known as:

(a) Scheduling, (b) Loading, (c) Routing, (d) Controlling.

Ans.(b)

71. One of the aims of loading is:

(a) To finish the job as early as possible, (b) To minimise the material utilisation, (c) To improve thequality of product, (d ) To keep operator idle time, material waiting time and ancillary machine timeat minimum.

Ans.(d)

72. One of the principles of Scheduling is:

(a) Principle of optimal product design, (b) Principle of selection of best material, (c) Principle ofoptimal operation sequence, (d) Principle of optimal cost.

Ans.(c)

73. The method used in scheduling a project is:

(a) A schedule of breakdown of orders, (b) Outline Master Programme, (c) PERT & CPM, (d) Schedulefor large and integrated work.

Ans.(c)

74. Production planning in the intermediate range of time is termed as:

(a) Production planning, (b) Long range production planning, (c) Scheduling, (d) Aggregate planning.

Ans.(d)

75. One of the requirements of Aggregate Planning is:

(a) Both output and sales should be expressed in a logical overall unit of measuring, (b) Appropriatetime period, (c) List of all resources available, (d) List of operations required.

Ans.(a)

76. In aggregate planning, one of the methods in modification of demand is:

(a) Differential Pricing, (b) Lay off of employees, (c) Over time working, (d) Sub contracting.

Ans.(a)

77. In aggregate planning one of the methods used to modification of supply is:

(a) Advertising and sales promotion, (b) Development of complimentary products, (c) Backlogging,(d) Hiring and lay off of employees depending on the situation.

Ans.(d)

78. The first stage of Production control is:

(a) Dispatching, (b) Scheduling, (c) Routing, (d) Triggering of production operations and observingthe progress and record the deviation.

Ans.(d)


299

79. The act of releasing the production documents to production department is known as:(a) Routing, (b) Scheduling, (c) Expediting, (d) Dispatching.

Ans.(d)

80. One of the important production documents is:

(a) Design sheet of the product, (b) List of materials, (c) Route card, (d) Control chart.

Ans.(c)

81. One of the important charts used in Programme control is:

(a) Material chart, (b) Gantt chart, (c) Route chart, (d) Inspection chart.

Ans.(b)

82. The way in which we can assess the efficiency of the production plant is by:

(a) Efficient dispatching, (b) By manufacturing a good product, (c) By comparing the actual performancewith targets specified in the specified programme, (d) By efficient production planning.

Ans.(c)

83. Production control concerned with:

(a) Passive assessment of plant performance, (b) Strict control on labours, (c) Good materialsmanagement, (d) Good product design.

Ans.(a)

84. Dispatching is toughest in:

(a) Job Production, (b) Mass production, (c) Batch production, (d) Flow production.

Ans.(c)

85. One of the documents to be prepared in dispatching department is:

(a) Time series for forecasting, (b) Cost sheets of products, (c) A shop order for each component,(d) None of the above.

Ans.(c)

86. The type of dispatching function to be used depends much on the:

(a) Type of product, (b) Type of production and on the size of the plant, (c) Type of materials used inproduction, (d) Type of management style.

Ans.(b)

87. One of the aims of dispatching is:

(a) To look after the welfare of the labour, (b) To test the quality of the product, (c) To dispatch theproducts to different places, (d) Principle of completion of job by due date.

Ans.(d)

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88. When work centers are used in optimal sequence to do the jobs, we can:

(a) Minimise the set up time, (b) Minimse operation time, (c) Minimise the break down of machines,(d) Minimise the utility of facility.

Ans.(a)

89. Route card and technological route card are:

(a) Different type of documents, (b) Same type of documents, (c) Route card shows route andtechnological route card shows the technology used, (d) One is prepared by production manager,dispatcher prepares other.

Ans.(b)

90. The card, which is prepared by dispatching department to book the labour involved in each operations:

(a) Labour card, (b) Wage card, (c) Credit card, (d) Job card

Ans.(d)

91. The card, which shows, the number of rejected products from total quantity produced is:

(a) Quality control card, (b) Inspection card, (c) Rejection card, (d) Job card.

Ans.(b)

92. The act of going round the production shop to note down the progress of work and feedback theinformation is known as:

(a) Follow up, (b) Dispatching, (c) Routing, (d) Trip card.

Ans.(a)

93. One of the activities of expediting is:

(a ) To file the orders in sequence, (b) To decide the sequence of operation, (c) To record the actualproduction against the scheduled production, (d) To examine the tools used in production.

Ans.(c)

94. The class timetable is a good example of:

(a) Route sheet, (b) Follow-up chart, (c) Dispatcher’s chart, (d) Gantt chart.

Ans.(d)

95. ‘Z’ chart is a chart used in:

(a) Programme control, (b) Job control, (c) Cost control,(d) Quality control.

Ans.(a)

96. Z-chart can be used to show:

(a) Process used in production, (b) Quality level of the product, (c) Both the plan and the performance,and deviation from the plan, (d) To show cost structure of the product.

Ans.(c)


301

97. Computers are used in Production control in this area:

(a) Follow-up activity, (b) To control labour, (c) To disseminate information, (d) Loading, Schedulingand Assignment works.

Ans.(d)

OTHER TYPES

Question

Match the Activity/Operation in Column I with the machine/equipment in Column II.

Answer

A: viii. Belt Conveyor

B: vii. Fork-lift Truck

C: vi. E.O.T. Crane

D: iii. Gravity Chute

E: iv. Drilling Machine

F: v. Planning Machine

G: ii. Electric Arc Furnace

H: i. Electromagnet

I II A. Feeding coal continuously into the furnace

in an Electric Power Station. i. Electromagnet

B. Handling crates on Pallets within a factory.

ii. Electric Arc Furnace

C. Moving a heavy load above the machine on the shop-floor in a workshop.

iii. Gravity Chute

D. Transporting fertiliser packed in bags to a railway wagon/truck on the ground below.

iv. Drilling Machine

E. Making a small deep hole in block of metal.

v. Plaining Machine

F. Machining a large flat surface on metal. vi. E.O.T. Crane G. Melting steel for making castings. vii. Fork-lift Truck H. Picking up bits of iron and steel in a scrap

yard. viii Belt Conveyor

302


Question

Match the product in Column I with the production centre/equipment/plant in Column II.

I II A. Furniture i. Assembly line B. Hydro-electricity ii. Refinery C. Television set iii. Foundry D. Cement iv. Carpentry E. Rails v. Smithy F. Aviation Fuel vi. Turbo-Alternator G. Tools vii. Blast Furnace H. Castings viii. Rotary Kiln 1. Forgings ix. Rolling Mills J. Pig Iron x. Machine shop

I II A. Furniture iv. Carpentry B. Hydro-electricity vi. Turbo-Alternator C. Television Set i. Assembly Line D. Cement viii. Rotary Kiln E. Rails ix. Rolling Mills F. Aviation Fuel ii. Refinery G. Tools x. Machine Shop H. Castings iii. Foundry I. Forgings v. Smithy J. Pig Iron vii. Blast Furnace

Answer

Matching:


303

Question

Match the terms shown under ‘X’ with their relevant terms; shown under ‘Y’.

X Y a. Ranking Method 1. Method Study b. Motion Economy 2. Plant Layout c. Work Sampling 3. Job Evaluation d. Normal Curve 4. Material Handling e. Use of Templates 5. Inventory Control f. Gravity Chute 6. Statistical Quality Control g. Crashing 7. Network Analysis h. Replacement 8. Value Analysis i. Brainstorming 9. Work Measurement

j. Stock Level 10. Maintenance

X Y a. Ranking Method 3. Job Evaluation b. Motion Economy 1. Method Study c. Work Sampling 9. Work Measurement d. Normal Curve 6. Statistical Quality Control e. Use of Templates 2. Plant Layout f. Gravity Chute 4. Material Handling g. Crashing 7. Net work Analysis h. Replacement 10.Maintenance i. Brainstorming 8. Value Analysis j. Stock Level 5. Inventory Control

I II (A) Steam (a) Blast Furnace (B) Electricity (b) Boiler (C) Steel (c) Generator (D) Petrol (d) Open Hearth Furnace (E) Iron (e) Refinery (F) Cloth (f) Assembly Line (G) Car (g) Smithy (H) Castings (h) Spinning Mill (I) Cotton Yarn (i) Foundry (J) Forgings (j) Power Loom

Answer

Matching:

Question

Match the products in Column I with the production centers in Column II.

304


Answer

Matching:

I II (A) Steam (b) Boiler (B) Electricity (c) Generator (C) Steel (d) Open Hearth Furnace (D) Petrol (e) Refinery (E) Iron (a) Blast Furnace (F) Cloth (j) Power Loom (G) Car (f) Assembly Line (H) Castings (i) Foundry (I) Cotton Yarn (h) Spinning Mill (J) Forgings (g) Smithy

Question

State whether the following statements are TRUE or FALSE.

(i) Method Study should precede Work Measurement.

(ii) Merit Rating is used to determine the cost of a product.

(iii) Production planning is an essential function in a factory.

(iv) Training boosts employee morale.

(v) A good Materials Handling system always consists of conveyors.

(vi) Increased productivity leads to cost reduction.

(vii) Project costs increase as the duration of the project increases.

(viii)When demand does not exist in the market, we should start Production Incentives.

(ix) A work stoppage generally reduces the cost of production.

(x) No handling is the best handling.

Answer

i. True ii. False iii. True iv. True v. False vi. True vii. True viii. False ix. False x. True.

Question


(a) It is desirable to conduct work measurement after Method study.

(b) Job Evaluation is used to measure absolute job worth.

(c) Incentive scheme is introduced by Management with a view to reduce direct labour cost.

(d) The increase in productivity can be attributed to the application of Industrial Engineering/Techniques,particularly the work study.


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(e) Operation process chart incorporates all five symbols.

(f) Multiple Activity chart deals with layout problems.

(g) Standard performance is the natural rate of working of an average operator when he works tinderproper supervision but without any financial motivation.

(h) Allowances for non-availability of materials power failure and breakdown of machines are providedfor in the standard time for an operation/job.

(i) In carrying-out Job Evaluation studies, point system is the best method.

(j) It is justified to consider the effect of working condition both in Work Measurement and Job-Evaluation.

Answer

(a) True (b) False (c) False (d) True (e) True (f) False (g) False (h) False (i) True (j) True.

Question


a. Adverse working condition would not be a factor in job evaluation where this is already compensatedby suitable allowances in the Standard Time.

b. Standard Time for a Job should be more in India than in Europe because labour productivity here islow.

c. It is a proven fact that higher productivity invariably results in poorer quality of products.

d. There is a limit beyond which labour productivity cannot be improved.

e. Standard Man Hour is the only unit that can be used for output measurement in all types of factories.

f. Job enrichment means increased work load for the worker.

g. Methods improvement has unlimited scope.

h. Value Engineering aims at reducing work content of a product.

i. Primary objectives of an incentive scheme should be to link a part of the workers’ earnings withproductivity.

j. Job Evaluation establishes relative job worth only.

Answer

(a) False (b) False (c) False (d) True (e) False (f) False (g) True (h) False (i) True (j) True.

Question


a. Increase in productivity leads to retrenchment of work force.

b. In view of rapid technological advancement we would not concentrate on labour productivity.

c. Piece wage system is a substitute for proper supervision.

d. Personnel Manager has nothing to do with productivity. It is the job of Technical Personnel.

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S. No. Metal working process Machine/Equipment i. Turning Centre Lathe ii. Drilling Pillar Drill or Radial Drill iii. Honing Honing Machine iv. Shaping Shaping Machine v. Milling Milling Machine vi Welding Welder (Gas or Electric)

e. Ranking is one of the Job Evaluation Techniques.

f. Results available from work sampling study is not 100% accurate.

g. Since breakdown of Plant and machineries is a random phenomenon, it is impossible to do any workmeasurement in Maintenance Area.

h. Job Evaluation does not help in performance Rating i. There is no difference between Method studyand Value Engineering.

i. Two-handed process chart is the most suitable Recording Technique in Electronics Assembly Industry.

Answer

(a) False (b) False (c) False (d) False (e) True (f) True (g) False (h) False (i) False.

Question

Mention the name of the metal working process involved in carrying out the following operations and alsostates the machine/equipment on which this is carried out.

(i) Reducing the diameter of a cylindrical object.

(ii) Making a cylindrical hole of an object.

(iii) Very line finishing of the inside diameter of a cylinder liner,

(iv) Cutting a ‘V’ groove on a flat surface.

(v) Reducing the thickness of one side of a metal cube.

(vi) Joining two metallic objects.

Answer

Note:- Alternative to vi. it can be done by Riveting also. In such case Pneumatic/or steam riveter can beused.

Question

Select general purpose machine tools to produce the following:

(i) ‘V’ groove in a Vee-block

(ii) Helicel groove on shaft

(iii) Thread in a nut


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(iv) Hole in a block

(v) Portion of shaft to be supported in a bearing sleeve

(vi) A flat surface on a large foundation block

(vii) A flat face at the end of a shaft

(viii)Teeth on a gear wheel.

Answer

S. No. Machine i. Shaping Machine ii. Milling Machine iii. Lathe iv. Drilling Machine v. Grinding Machine vi. Planning machine vii. Lathe viii. Milling machine or Hobbing machine

Question

Mention 8 different techniques which are used for improving productivity in industry.

Answer

1. Method study.

2. Motion and time study.

3. Ergonomics (or Human Engineering).

4. Network Analysis-PERT/CPM etc. for planning.

5. Value Analysis.

6. Statistical Quality Control (SQC).

7. Operation Research-Linear Programming etc.

8. Inventory control.

9. Budgetary control.

10. Management by objectives (MBO).

Question

What recording technique will you use for the following production situation ?

i. To have bird’s view of a chemical process.

ii. To develop work bench layout .

iii. To reduce movement of materials.

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iv. To study one operator running one machine.

v. To reduce ineffective body movements.

Answer

S. No. Recording Technique i. Outline Process Chart ii. Travel chart iii. Flow Diagram iv. Machine Data Card v.

First Motion study Second Man-Machine chart

Question

Give your views on the following statements:

a. Incentives are substitute for lower wages.

b. Mechanisation and Automation lower employee morale.

c. Total productivity of a given situation cannot be measured in absolute terms.

d. Method study should precede work measurement.

Answer

a. No. Incentives are in addition to wages and are meant for increasing production and improvingproductivity.

b. No. They will simplify job and as such morale will improve.

c. No. In such case output and input can be converted into monetary terms and productivity measured.

d. Yes.

Question

Give your views on the following statements:

a. Incentives are substitute for lower wages.

b. Job Enlargement (and not the division of labour) is the key to higher productivity.

c. Method study should precede work measurement.

d. A standard Job Evaluation Manual can be applied to any group of jobs for the purpose of Job Evaluation.

Answer

a. No Incentives are in addition to wages and are meant for increasing production and improvingproductivity.

b. Yes.

c. Yes. Methods should be standardised and then work measurement (i.e. Time study) is to be undertaken.

d. No.


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Question

List six resources on which quality depends.

Answer

(1) Shape (2) Dimension (3) Composition (4) Strength (5) Workmanship (6) Finish.

Question

Answer the following queries with ‘Yes’ or ‘No’.

(i) Would Zero-based budgeting require greater involvement of operating departments?

(ii) Is electric power required to run a gravity roller conveyor?

(iii) Does the time over-run in a project affect its cost?

(iv) Do Job-Evaluation and Value Analysis concern the same technique?

(v) Does the work ‘tolerance’ in engineering practice refer to the ability of workmen to endure boredom?

Answer

(i) Yes (ii) No (iii) Yes (iv) No (v) No.

Question

Match the material handling equipment in column I with the application in column II.

I II A. Fork lift Truck (a) move bulk material continuously B. Jib crane (b) move heavy loads over rectangular area C. Belt Conveyor (c) move heavy loads within a circular area D. Electric Overhead Travelling crane (d) move loads down E. Gravity Chute (e) move palletised unit loads

r Matching Particular A — e Fork lift Trucks move Palletised unit loads. B — a Jib cranes move bulk material continuously. C — c Belt Conveyors move heavy loads within a circula area. D — b Electric Overhead Travelling (EOT) cranes move heavy loads over

rectangular area. E — d Gravity Chutes move loads down.

Question

Indicate whether the following statements are TRUE or FALSE:

(i) A bill of materials is prepared by the Sales Department in an organisation.

(ii) Progressing ensures that production takes place according to plan.

(iii) Most castings are made in green sand mouldings.

Answer

310


(iv) Only railway workshops use engine lathes.

(v) Drawings, where applicable, are the best: means of expressing product specification.

(vi) Work content of a job is established by Job Evaluation.

(vii) Industrial Engineering is not a line function.

(viii)Scanlon plan is a system of production, planning and control.

Answer

(i) False. (v) True.

(ii) True. (vi) False.

(iii) True. (vii) True.

(iv) False. (viii) False.

Question

Answer the following Queries:

(i) Do standard Times allow for relaxation of the Operators?

(ii) Is a lift same as an elevator?

(iii) Is the use of metric system of weights and measures compulsory in India?

(iv) Can the shaping machine be considered a versatile machine tool?

(v) Is payment of Gratuity to Workmen compulsory in India now?

(vi) Does the Factories Act in India allow the employment of women in all industries?

(vii) Is Break-even analysis a management tool?

(viii)Is Activity Sampling a technique of Job Evaluation?

Answer

(i) Yes (v) Yes

(ii) Yes (vi) No

(iii) Yes (vii) Yes

(iv) No (viii) No


311

Question

Given below are two lists—list ‘A’ containing 11 abbreviations and list ‘B’ containing various functionalareas associated with production management. Expand the abbreviations and match them with thecorresponding functional areas.

List ‘A’ List ‘B’ LP Capacity planning PERT Quality control MTM Project funding VA Project viability checking CRAFT Inventory management SRAC Product design MRP Cost control CBA Product mix determination CAD Plant layout IFCI Project planning AOQ Work measurement

Answer

List ‘A’ Expansion Matching with List ‘B’ LP Linear Programming Product mix determination PERT Programme Evaluation and Review Technique Project planning MTM Methods Time measurement Work measurement VA Value Analysis Cost control CRAFT Computerised Relative Allocation of Facilities

Technique Plant layout

SRAC Short Run Average Cost Capacity planning MRP Materials Requirement Inventory management CBA Cost Benefit Analysis Project viability checking CAD Computer Aided Design Product design IFCI Industrial Finance Corporation of India Project funding AOQ Average outgoing Quality Quality control

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X Y (i) Foundry (a) Gearbox (ii) Machine shop (b) Value Analysis (iii) Brainstorming (c) Electrode (iv) Automobile (d) Lathe (v) Forge Shop (e) Cupola (vi) Welding (f) Power Hammer (vii) Heat Treatment (g) Rubber (viii) Tyre Plant (h) Hardening (ix) Assembly line (i) Go-No Go gauge (x) Inspection1 (j) Conveyor

X Y

(i) Foundry (e) Cupola (ii) Machine Shop (d) Lathe (iii) Brainstorming (b) Value Analysis (iv) Automobile (a) Gearbox (v) Forge Shop (f) Power Hammer (vi) Welding (c) Electrode (vii) Heat Treatment (h) Hardening (viii) Tyre Plant (g) Rubber (ix) Assembly Line (j) Conveyor (x) Inspection (i) Go-No Go Gauge

I II (a) UCL (i) Work Measurement (b) MTM (ii) Project Management (c) CPM (iii) Public Sector (d) LP (iv) Quality Control (e) BIS (v) Optimisation (f) FIFO (vi) Standardisation (g) MTBF (vii) Reliability (h) BPE (viii) Inventory Management (i) WIP (ix) Cost Accounting (j) VOH (x) Production Control

Question

Match the terms shown under ‘X’ with the relevant terms shown under ‘Y’.

Answer

Question

Match the terms in column I with the relevant terms in column II.


313

Answer

(a) UCL (iv) Quality Control (b) MTM (i) Work Measurement (c) CPM (ii) Project Management (d) LP (v) Optimisation (e) BIS (vi) Standardisation (f) FIFO (viii) Inventory Management (g) MTBF (vii) Reliability (h) BPE (iii) Public Sector (i) WIP (x) Production Control (j) VOH (ix) Cost Accounting

Question

Given below are two lists-List A containing abbreviations and List B containing various functional areasassociated with production management. Expand the abbreviations and match them with the correspondingfunctional areas.

List A List B LP Product design PERT Quality control MTM Project funding VA Project viability checking CRAFT Inventory management MRP Cost control CBA Product-mix determination CAD Plant layout IFCI Project Planning AOQ Work measurement

Answer

(i) LP — Linear Programming — Product mix determination.

(ii) PERT — Programme Evaluation and Review Technique — Project planning.

(iii) MTM — Methods Time Measurement — Work measurement.

(iv) VA — Value Analysis — Cost control.

(v) CRAFT — Computerised Relative Allocation of Facilities Techniques — Plant layout.

(vi) MRP — Materials Requirement Planning — Inventory management.

(vii) CBA — Cost Benefits Analysis — Project Viability checking.

(viii)CAD — Computer Aided (or Assisted) Design — Product Design.

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I II (a) ISO (i) Work Study (b) PBT (ii) Inventory Control (c) CNC (iii) Standardisation (d) JIT (iv) Profitability (e) PMTS (v) Machine Tool (f) ESI (vi) Computer Programme (g) COBOL (vii) Public Sector (h) ITI (viii) Welfare (i) CPM (ix) Employee Relations (j) IR (x) Project Management Answer (a) ISO (iii) Standardisation (b) PBT (iv) Profitability (c) CNC (v) Machine Tool (d) JIT (ii) Inventory Control (e) PMTS (i) Work Study (f) ESI (viii) Welfare (g) COBOL (vi) Computer Programme (h) ITI (vii) Public Sector (i) CPM (x) Project Management (j) IR (ix) Employee Relations

(xi) IFCI — Industrial Finance Corporation of India — Project funding.

(x) AOQ — Average outgoing Quality — Quality control.

Question

Match the terms in column 1 with the relevant terms in column II.

Answer


315

Question

Expand the following 10 abbreviations indicated in column X and then match the same with the mostappropriate one indicated in Column Y on the right-hand side.

X Y SPT Standardisation ICICI Labour related standards ABC Scheduling ISO Tax based on cost of additional processing PPC Venture Capital LCL Machines used for producing a class of products SPM Manufacturing planning and monitoring VAT Marketing strategy USP Statistical Quality Control ILO Classification based on annual usage value Answer SPT Shortest Processing Time — Scheduling ICICI Industrial Credit and Investment Corporation of India — Venture

Capital ABC Always Better Control — Classification based on annual usage value ISO International Standards Organisation — Standardisation PPC Production Planning and Control — Manufacturing planning and

monitoring LCL Lower Control Limit — Statistical Quality Control SPM Special Purpose Machine Tools — Machines for producing a class of

products VAT Value Added Tax — Tax based on additional Cost of processing USP Unique Selling Proposition — Marketing strategy ILO International Labour Organization — Labour related standards

Question

Choose the word or phrase which would be appropriate to fill up the blanks in each statement:

(i) Statistical analysis is used to determine the optimum policy of ________maintenance.

(ii) Watch and ward personnel are responsible for ________aspects in a factory.

(iii) General purpose machine are less prone to___________.

(iv) The pattern shop in a factory should ideally be near the____________.

(v) Factor Comparison is a method of_____________.

(vi) Taylor originated the idea of__________relationships in an organisation.

(vii) _____________cannot be delegated.

(viii)Ergonomics is another name for_________________.

(ix) _______________layout is used for mass production.

Answer

316


I II (A) Foundry (a) Gauge (B) Smithy (b) Conveyor (C) Machine shop (c) Mould (D) Welding (d) Forge (E) Heat treatment (e) Lathe (F) Assembly line (f) Microscope (G) Inspection (g) Stacker (H) Laboratory (h) Furnace (I) Design office (i) Electrode (J) Warehouse (j) Pantograph Answer

I II (A) Foundry (c) Mould (B) Smithy (d) Forge (C) Machine shop (e) Lathe (D) Welding (i) Electrode (E) Heat treatment. (h) Furnace (F) Assembly line (b) Conveyor (G) Inspection (a) Gauge (H) Laboratory (1) Microscope (I) Design office (j) Pantograph (J) Warehouse (g) Stacker

(x) Gantt chart is used for_____________ control.

Answer

(i) preventive

(ii) security

(iii) obsolescence

(iv) foundry

(v) job evaluation

(vi) functional

(vii) responsibility

(viii)human engineering

(ix) product

(x) production

Question

Match each of the words in column I with the most appropriate ones from column II:

➏➐

Answer

Part - IIInformation

System

InformationSystem

STUDY NOTE - 1

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1.1 Contribution of Information Technology in the Social life and Industrial fields

Information Technology has become the most dynamic area in human life today. In social life, fromeducation, entertainment to economic development programme, everywhere information processing givesdirection for plans and programmes. The world has become a global village with the power ofcommunication technology. Physical location is no barrier for communication. Power of computers andcommunication technology has redefined information technology and has brought a new dimension in itsuse. A business can not run with information flow. The visualization of importance of timely informationhas created new atmosphere for intensive research and development programme for evolving newtechniques and tools in the field of information technology. This is why we call that the present age is theage of Information.

We see the marvelous power of computer and it has grabbed all areas of modern life. Its speed, accuracyand storage facility have charmed everybody. The inherent aptitude of man for mathematics felt the needfor a machine to calculate the arithmetical figures. With this urge man could develop a device which hasbrought a new taste of life. Innovations in the use information technology are examples of human brainpower. No body can guess the end of it. A small list can only give an amazing dimension - use of robot tocontrol production and quality in the factory, conference among people at different corners in the worldthrough communication, remote control of space craft from Earth Station and so on. Nothing is left onearth which is untouched by IT today. Each and every activity of day–to-day life like to have a goldentouch of it. Let us have the glimpses of its small range which are

• Education and Training

• Medical science

• Engineering science

• Space research

• Research in bio-technology

• Entertainment etc.

Information is the backbone in a business organization. In today’s competition, managing information isthe most critical task for survival. Information Technology plays a crucial role in dissemination ofinformation towards controlling business operation. The work of computer has been extended from dataprocessing to on-line Management Information system, Sales information to E-commerce, Decision makingto Expert system. The challenges ahead are the fields like:

• Process Control in production management

• Flow of information through internet

• Integration of functional systems with network

• E-Commerce in business

• Knowledge based Expert System in decision making etc

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1.2 Information

Information is nothing but refined data. Data before processing is said to be raw data. Raw data are collected,screened and processed to make it organized for effective use. Data after processing become linked withother data and carry meanings and, strictly speaking, to be termed as information. Information is data thathas been processed into a form that is meaningful to the user in effective interpretation and decision making.Information involves communication and reception of intelligence. Information consists of data, text, images,voice etc. The term data in normal sense includes all these.

Processing

Information

DecisionTools

Decision

Data

SkillKnowledgeExperience

Characteristics of useful Information:

The following are the characteristics of information. These characteristics determine the quality andusefulness of information :

• Timeliness – This parameter is important to increase effectiveness in the use.

• Accuracy – The most important ingredient for quality of information.

• Comprehensive – Information should be integrated one with all other related issues to make it moremeaningful.

• Relevance - The need for type of information differs from user to user. Relevant information filteredfor a purpose ensures its effective and best use.

• Understandability - The information must be presented in a form that users can interpret the samefor decision making.

1.3 Information System

An Information System is designed by an user according to the business operation involved and the needfor his decision making. It is a vehicle which supplies necessary information for decision making. Beforecomputer being used in an business environment, Information System also existed in some form to supportmanagement in decision making. To-day business environment has become very complex and highlycompetitive. If an organization has to survive, it has to plan and develop its own Information System. Ofcourse, the size, structure and set up may vary depending upon the need of the organization, policy andcapability. In today’s business environment computer has become a must for an information system. The

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information system is to feed the management for control purpose. This is why Information Systemis generally called as Management Information System (MIS) or Computer Based Information System(CBIS).

Generally, am MIS uses Computer System and Communication technology to collect information fromdifferent operational points and disseminate them at different users for decision making. The activities inan Information System are :

• Collection, storing and processing data

• Generation of Information Reports

• Dissemination of Information to right users

An MIS provides timely information of right quality for better management decision making for developingbusiness strategy. The advantages of CBIS are :

• Reduction in cost of record maintenance

• Improvement in the efficiency of human resources

• Regular flow of information at different levels of management

• Easy use of scientific tools and models for quality decision making

• Faster response to customers

• Better control over resources

• Faster access to records in case of dispute

• Effective use of manpower etc.

Characteristics of an Information System : - The following are the general characteristics of an Informationsystem :

i) Specific objective : The information system should have some specific objective. An Information Systemin highly scientific research centre will have an objective to accumulate data from different activities,display of some information instantly for controlling activities and so on. In a business environment,the objective will be sharing information from different functional areas and smooth flow of informationfor management decision making.

ii) Structured : An information system should have a definite structure with all modules of sub-systems.The structure depends on the sub-modules, their interactions and integration requirements, operationalprocedure to be followed and the solution sets. The structure of the information system refers todiagrammatic representation of the system showing sub-systems, their inter-relation and the procedureto be followed to fulfill the process requirements.

iii) Components : The sub-systems are the components. The sub-systems should be distinguishable amongthemselves but have well-defined relation among. For example, a Sales system may be sub-systemslike Invoicing, Delivery monitoring, and Sales Proceeds Collection system. The inter-link betweenthese systems must be well defined.

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iv) Integrated: An Information System should be designed in such a fashion that proper integration amongsub-systems are taken care to establish correct linkage and generate meaningful information. Aninformation in isolation may not be that meaningful but its usage is improved if it is integrated withinformation of other closely related issues. For example, Sales information of a region becomes moremeaningful if other information like previous period sales, sales in other regions, sales of competitiveproducts are also combined in the information set.

v) Life-Cycle: An Information system will have its own life-cycle. The duration of life cycle varies fromthe system to system. An information system has the similar stages of life-cycle as seen in any othersystem. Every information system will have distinctly different phases - Initial, Growth, Maturityand Decline.

vi) Behaviour: A system has its own set reaction and outcome depending on the environment. A wellmanaged business information system behaves nicely with its users by satisfying them with correctand timely information. The design of the system plays a good role in setting its behaviour pattern.

vii) Self-regulatory: An Information System which may have different sub-systems interacting with theeach other in a desired fashion to be operative smoothly and in the process they regulate themselves.This is what is self-regulatory nature of the system. A payroll system involves three activities – first,maintaining attendance of employees, second, pay calculation and third pay disbursement. If thetarget date for pay disbursement is last date of a month, the second adjusts its start time accordinglyand the first one is also regulated in such a fashion that it can provide input to the second in time.

1.4 Open and Closed System

1.4.1 Open System: An open system is one which interacts with its environment and can change itself toaccommodate the changes in factors like customers’ preference, price, product design etc. The adoptabilityof an open system is judged by its capability in modifying the operational parameters of the systemaccordingly. It takes input from outside world and exports output to outside world. For example, a FinancialManagement system can function smoothly even with the change in credit period and credit limit from thebanker is said to be having the flexibility to adjust itself with the change the environmental parameters. Ifa system can accommodate all the changes in the environmental factors, as and when required, is said tobe perfectly open system.

1.4.2 Closed System: A closed system which does not have any interaction with outside environment. Inother words, it functions in the closed environment set and is insular with the change in the environment.A closed system is a self contained one and normally a rigid one. As business environment is subject tochange and a business system is expected to adjust itself with the change in the environmental factors. Inother words, a business system can not be closed system. In fact, close systems are very limited. Somesystems in Military or Defence Service may be closed system because their rules, procedure and factors areset to be rigid for the sake of strict code of discipline.

1.5 Evolution of Information System

The evolutions in Information Systems can be distinctly divided into five Generations of Information Systemwhich are as follows :

1. Manual System

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322

2. Mechanical System

3. Electronic Data Processing (EDP)

4. Management Information System (MIS)

5. Decision Support System (DSS)

6. Executive Information System (EIS)

7. Expert System

1.5.1 Manual System (before 1930)

• It was old and primitive

• It involved highly clerical work and chance of procedural errors was high

• Procedure was time consuming and there used to be delay in preparation of reports

• There was no alternative cost-effective procedure

1.5.2 Mechanical System (1930-1955)

• Unit Record Machines came into existence

• Improved Computational accuracy

• Comparative faster data processing

1.5.3. Electronic Data Processing (EDP) System ( 1955-1970)

• Focus was on volume-oriented applications

• Use of computers improved processing efficiency

• Well defined processing rules

• Development of scientific programming and system designing

In 1979, Richard Nolan suggested a model for evolution of information system which has been widelyaccepted by academicians and practitioners. The model suggest that the level of data processing expenditureof an organization can be seen to progress through six different stages which are as follows :

Stage Activities

Initiation The First introduction of Computer System was for cost saving.

The job which were repetitive in nature and needed high manpowerdeployment used to be done in computer.

Use of computers restricted to accounting function.

Knowledge became barrier for future development.

Contagion The blossoming of computer application in different areas came in a totallyuncontrolled way.

This stage became chaotic and many application fail.

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Stage Activities

Control Senior Managers became conscious about the expenditure and lack ofcontrol.

Staff are centralized and formal Information system organization started tobe established.

Application concentrated on saving money, rather than making money.

Integration The control introduced in stage 3 is slackened in order to introduceinnovation.

The Information system function was re-organized to allow the specialiststo become involved with users in development of systems.

Large expenditure was made in core system.

Data Administration Developments were driven by the organization’s demand for information.

The business recognized the value and potential of information system.

Corporate databases creation started.

Maturity Planning and organization in Information System got incorporated intodevelopment of the organization.

Strategy was embedded in decision making.

1.5.4 Management Information System – MIS has an objective to provide best possible timely informationwith the use of modern sophisticated technology. The strategy in MIS is to exploit the technological tool inensuring the flow of information of right quality at right time at right place (user’s place) for effectivedecision making. Its important features are :

• It is an improved version of EDP system

• It enables the management to have access to desired set of data only

• It meets information requirement at different levels with pre-defined reports

• It undertake transaction processing for different functional areas

• It provides on-line access to data and efficient reporting system

• It supports routine decision making

1.5.5 Decision Support System – It is a sophisticated decision making model with the help of high poweredsoftware to take semi-structured decisions. The basic features of a DSS are :

• It is based on one or more corporate databases

• It must be supported by a set of quantitative models

• It has the ability to solve unstructured problems

• It should have the network computing facilities embedded in the system

• It is used for solution in a complex business situation

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• It provides supports to Executive Information system for decision making

• Software at different locations facilitates group decision making.

1.5.6 Executive Information System : An Executive Information System is a advanced model of DecisionSupport System which can take care of unstructured problem situation. It aims at providing informationto top executives of an organization who are involved in strategic decision making.

1.5.7 Expert System : An Expert system is a knowledge based system which acts an expert in devisingsolutions. An expert system acts in a specific area only with the support of knowledge database on thisspecific area. Knowledge data base means structured information stored on previous solution sets inunstructured problem situations. In other words, an expert system operates on previous experience whichis stored in a database. Even the present solution devised from the system and the information on itsoutcome will also be stored.

1.6 Growth of Business Information System

In recent times there has been substantial growth for scientific information system in business system.Modern business managers have perceived the importance of support from good information system inbusiness decision making. The following factors have contributed towards growth of business informationsystem :

1.6.1 Change in Computer Technology: Cost of hardware has gone down. Thanks goes to technologicalinnovations. Computing speed has gone up tremendously and storage capacity has increased enormously.The revolution in communication technology induced every activity in the business and society to beunder its call. The change in the computing capabilities in Information systems due to technologicalinnovations have brought significant changes in the attitude of the management towards effective use ofinformation.

1.6.2 Fast Transaction Processing: Today network technology has brought the concept of intranet in anorganization. This has helped fast and accurate transaction processing based on instant updation ofinformation from point-of-sale ( POS) and warehouse. Large Banks are using the intranet for having on-line information for exchange and control. Information system in banking service has brought a newdimension in control, efficiency, productivity and profitability.

1.6.3 Customer Support: Market place is full of suppliers. Products are flooded from domestic andinternational supply. The expectations of the customers has increased. They have become moreknowledgeable about the quality, cost and availability of the products. A business organization can notsurvive unless it demonstrates its customer orientation. There has been a trend to manage the total supplychain system through computer system primarily to satisfy customer need and to efficiently controlinventory.

1.6.4 Changing Product Technology: In the global competition of trade, the technology is fast changing.Innovations are revolutionary. Products have shorter life-cycle. Risk of business obsolesce is very high.Only a good Information system can provide right kind of support with right flow of information forscientific control mechanism and decision making. Business Managers are under tremendous pressure tocope up with the speed of development. The magic tools under IT provide continuous information toupdate with the changes taking place throughout the world.

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1.6.5 Continuous Improvement Effort : Cost Competitiveness and quality improvement are two importantaspect on which survival of an business depends. Only continuous efforts with better control mechanismcan help to achieve the results. The information system is providing the best support. Is it possible to adoptJust-in Time for Inventory Control without computer based information system? Quality information withthe help of computer provides the right kind of database for analysis and corrective measures.

1.6.6 Strategic Decision Making : Business Strategy and Information system of a modern organization isinseparable. A sound strategic decision is based on analysis of various factors – both internal and external.Strategic Analysis model of Michel Porter can provide strategic solutions to business problems based onscientific and comprehensive information data base. The Model is an eye opener to the business world todevelop a new generation of information system for spontaneous strategic business solutions.

1.6.7 Wide Areas of applications in Offices : Computers are being widely used for different purpose inbusiness environment to support the cause of business. They are :

• Text management

• E-mail

• Document management with the use scanner

• Communication through network etc.

1.7 Strategic Planning for Development of an Information System

An Information System is developed with a definite set of objectives and plan to achieve them. Fordevelopment of a better system, change in the management system in regards to accountability,responsibility and control in different functional areas related to system is essential for effective flow ofinformation, decision making and monitoring.

Until most recently, most organizations regarded their Information systems as a resource that is necessarybut not strategically significant. The IT department was treated like any other service department and theinformation system was allowed to evolve rather than being formally planned. Major benefits for formalinformation system strategy :

• To achieve goal congruence between information system objectives and corporate objectives

• To create and sustain competitive advantage

• To exploit Information Technology to the optimum benefit of the organization

1.7.1 Linking Corporate Strategy with Information System Strategy

The basic objective of information system strategy is to exploit IT to provide best advantage for theorganization.

The following points are to clearly understood before venturing into designing an information system :

1. Identification of sub-systems involved and their interactions

2. Level of Management

3. Decision making process

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Today, success of corporate strategy depends on decision making skill. The support of efficient informationsystem needs no clarification. The decisions at different levels of management vary. A standard decisionmaking pattern is shown below :

Level of Management Decision making on Information Support from

Lower Level Operational Control Transaction Processing

Middle Level Planning & Control Management Information System

Top Level Strategic Planning and Executive SystemImplementation Expert System

Information and data architecture :

What is needed is to understand the following :

• Information need for the organization for corporate decision making

• Who will provide the data and who will use it.

• When and how data will be collected.

• How should the data be stored to provide easy access.

Application Architecture

Application architecture development requires the evaluation of the application software requirement inthe consideration with the following :

• What application software is required

• What is the most logical sequence of operations

1.7.2 Budgeting and Planning the Implementation details

To plan for future information system the following issues should be critically evaluated :

• Organizations growth plan, business strategy and support needed from Information system

• Information System Budget

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• Quality of Human Resources and possible approach to improve their skill

• Applicability of business process re-engineering

• Integration of information from different functional areas

1.7.3 Planning the Operation

To make the information system more effective, the action plan has to be devised for :

• Development of organizational structure in IT department

• Fixation of new role and responsibilities

• Security management

• Risk Management

1.7.4 How to Support the Decision making Process

The important process in the planning is how to improve the use of information in efficient decision makingto formulate business strategy. For doing this, through review of the current system and procedure forInformation Management is essential which involves the review of the following :

• Current work volume

• Current application software and their life cycles

• Current technology environment and skill level

• Current performance of Information system

• User satisfaction

• Current Business Process and information structure

1.8 Information System Infrastructure

Information System Infrastructure means the physical resources and organizational support required foroperation of an information system. It consists of following six basic components:

• Hardware – Devices which store software, database and process data

• Software – Programs process data to generate reports

• Database – Data collected is stored in databases

• Network – Technology for sharing the data and other hardware resources

• People – Human resources to make the system operational.

Reports – Reports are generated by the software with the help of databases for the use by users (people).

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The role of information system architecture is to support and reinforce the organization structure anddecision making mechanism. If we accept that IT can change the performance of an organization, the realchallenge before the management is to ensure the management that the components of IS architectureprovides the most suitable solution in the environment under which an organization works. The changemanagement is the most critical issue and change in the architecture should be taken up well before thetechnology becomes obsolete.

The following exercises are to be made for effective use of resources with the use of technology:

1.8.1 Arranging required hardware and software

Management has to ensure the most suitable hardware and suitable to fulfill the information requirementfor decision making. The choice between centralized processing or distributed processing has to be madefirst. The architecture design is dominated by issue of compatibility of hardware platform and softwarepackage.

• Assessment of hardware configuration on the basis of volume of data and type of processing need

• Software requirement in compatible with hardware

• Networking and communication technology requirement

• Assessment of investment requirement and phasing the investment

• Vendor selection

• Procurement plan

1.8.2 Human Resource Management

The system developed by IT specialist often fail to meet the user need. The persons who will be in-chargeof development of software must have clear understanding of how the business operates so that technically-perfect solution is evolved. What is needed is personnel involved in Information system must be properlytrained on the pattern of functioning of the organization, change needed, plan for changes, stages oftransformation and actions plan for the same. The policies of the relevant issues are to be framed to achievethe objectives. The issues are :

• Recruitment

• Training

• Restructuring

• Retention of employees etc.

1.8.3 Implementation

Implementation of infrastructure requires project management skill. What is required is to develop timetargets for implementation activities like procurement of hardware, software, allocation of resources andimplementation of the system. The time frame may vary from organization to organization forimplementation and level of sophistication depends on present system, possibility of skill improvementand acceptance in change in management of the system. Strategies are to be developed for achieving a timetargeted system with desired effectiveness. Careful considerations should be taken about the followingcriteria of an dynamic information system :

• Information needs of an organization change constantly

• The information system should be open and adaptive

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• The information system should focus on supportive role in the business process

1.9 Information System Organization

Organization structure should be based on established policy and have well defined rules of responsibilitiesand authority at different levels. The load of data processing, resource requirement in terms of bothmanpower and machine must be assessed properly. Job specifications at different levels must be clearlygiven to avoid gaps in responsibility and performance standard also be rationally established to make theorganization more scientific. The objectives of sound organization structure are to provide all possibleinfrastructure facilities for a good information system. To be more specific, a scientific organization structurefor information system means provisions of the following :

• Proper Information Technology environment with right kind of machine, manpower and work culture

• Right resources balancing the hardware, software and skill

• Adequate security system on data, processing and output

• Adoption of scientific and modern software development methodology etc.

For successful implementation and operation of an information system, organizational set up has to bebuilt to take care certain activities on a day to day basis. Let us understand the activities involved in theinformation system and the responsibilities of the Information System Department.

The organizational structure of information system in an organization may vary depending on variousresources available and their quality but the objectives remain almost same with main focus on effectiveuse of information for better business control.

Activities involved in the department are :

1. System Development

2. Programming

3. Data administration

4. Security management

5. Operation management

6. Quality assurance

The different specialist groups of employees are assigned the responsibilities of the above activities in theInformation System Department. The chart below will show the organization structure of a Computerbased Information System department :

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The following table will explain their activities and responsibilities:

Designation Duties and responsibilities

Information System Manager • Planning the resources and time frame of implementation

• Supervising the overall implementation of system and dayto day operation

Database Administrator • Database Management

• Database Library Management

• Security of data

Business Analyst • Development of new Information System.

• Acting as co-ordinator between users and IS developers

Systems Analyst • Analysis of the requirements of the IS, development ofsystem specification

• Designing the IS.

Programmer • Development / modification of programs according tosystems specifications and design.

Maintenance Engineer • Maintenance of Hardware

Network Specialist • Maintenance of network

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Designation Duties and responsibilities

Systems Operation Manager • Planning and controlling the flow of data and processing.

• Network management.

• Co-ordination with users.

Operator • Computer Operations

• Communication network control

• Process Documentation

• Data Entry Supervision

Data Entry Operator • Data Entry, Data verification and editing

1.10 Information Systems Personnel Management

Managing the human resources effectively is very critical issue. Success of a Information Sytem dependson the quality of human resources and their support.

The turnover in Information Technology field is comparatively high compared to other industry. Thus,following important issues should be taken into :

• Growth prospect

• Motivational aspects

• Provision for management of gap in skill and expertise

For effective management of human resources, the following policy and procedure must be followed :

1. Job description for different positions should be clearly spelt out to avoid conflict and confusion.

2. Job responsibility for each position should be clearly defined.

3. Recruitment policy must be well defined one to hire right quality personnel.

4. Training programs should aim at skill improvement of the personnel in line with technological changeand organization’s requirement.

5. Regular screening of security check must be followed to plug all possible motive for fraudulent activitiesand to avoid risk of any damage.

6. Performance evaluation should be rational, scientific and unbiased to motivate the employees.

1.11 Quality of a Business Information System

Main quality criteria of a business information system are reliability of information, timeliness andcorrectness of reports. These depend on technology in use, manpower, ethical standard maintained andsecurity of the system. To ensure good quality, system requirements must be properly understood andsystem infrastructure must be developed.

1.11.1 Reliability of Information : Internet has flooded the information. Today business managers areburdened with analysis of huge volume of information. The exact of relevant information only can lead to

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sensible use of information. The role of Business Analyst will be of paramount importance with the growthof technological evolutions. There is a saying in computer information system – Garbage In Garbage Out(GIGO). The correctness or quality of reports depends on quality of information. Source of correct data,flow of data and procedure of authorization has to be established.

1.11.2 Processing Time : Processing time is an important factor for information system. Information isgenerated for control in different business parameters in the operation and decision making. Naturally,the time associated with dissemination of information is very important. Information not in time becomesuseless.

1.11.3 Matching Management Requirements : The information system should be well planned to fulfill therequirement of the management i.e. the owner of the system in terms of time, quality and frequency. At thetime of designing the system the reports, contents, formats etc. are to taken care to exactly fulfill therequirements of users of different levels.

1.11.4 Technology : Technology refers to :

• Configuration of computer in terms of capability, security etc

• Software in use

• Communication system – network efficiency

• Type of processing - batch, real-time or on-line

1.11.5 Human Resources : The employees need new orientation in the knowledge of computer products,technological change taking place, expertise to handle these and managerial complexity evolving out ofthese. They are to develop the skill to develop appropriate IT infrastructure, use effective methods to storeand access data and navigate them to properly use them. Quality of human resources is the most importantcomponent in an Information System organization.

1.11.6 Ethical standard : Ethical issues relating to employees, customer, supplier etc must be properlydealt with to maintain the right kind of image of the organization. Ethical standard must be maintained indealing with the information being handled in an organization.

1.11.7 Security : The security of information system must be taken care properly. Here, security meanssecurity of information and assets. In an processing environment under network, the system is morevulnerable by unauthorized assess, leakage of sensitive information and sabotage. Loss of assets due tofire, theft, natural disaster must be protected.

1.12 Business opportunities due to Development of Computers

Development in Computer has also evolved following business opportunities for Different Business activitiesrelating to Hardware and Software.

1. Computer Manufacturing : This a function generally taken up by big companies. Earlier daysmanufacturers used to market their own products. Leading manufacturers in the world were IBM(International Business Machine), DEC (Digital Equipment Corporation), CDC (Control DataCorporation), Burrough, Honeywell, NCR, ICL, Hewlett Packard etc.

2. Peripheral Equipment Manufacturing : Some companies are involved in manufacturing peripheraldevices like disk drives, tape drives, printers, plotters, monitors, communication equipments. Leading

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peripheral manufacturers in the world are Intel Corporation, Motorola, Texus Instruments, Toshiba,Phillips, Fujitsu, Hitachi, NEC Corporation etc.

3. Computer Leasing: Earlier days Computer Leasing was a very attractive business. The cost of Computerwas high. Users generally were not ready to afford huge investment. Some computer manufacturerand some companies were involved in the business of computers leasing. Lease Rent was high. IBMused to provide machines in lease.

4. Software Developers: Companies and individuals undertake the responsibilities of developing softwarefor third party companies. The job of contract used to be system designing, program developmentand system implementation.

5. Time-Sharing Companies: In earlier days when hardware used to be costly and users were using onlyselective application systems for data processing, the time sharing facilities provided by differentcompanies on their Computer were found to be cost-efficient and widely acceptable to many users.Some Companies took the opportunities for business purpose. Generally, companies used to buy amainframe computers with multi-users facilities and charges used to on the basis of usage time ofdifferent components like CPU, Printers, Disk, Magnetic tape etc.

6. Networking Parlour : With the acceptance of networking to be the most efficient communication, useof e-mail and net-working surfing has gone up. It will further multiply when the e-commerce willtake a formidable shape to be effectively operative. Providing networking facilities by small shop hasbecome a good business in big and small towns.

7. BPO Services : The advantages of fast communication though internet technology has opened a newdimension for Business Process Outsourcing of commercial jobs. The data in being transmitted fromone part of the globe to other for processing. The same being processed.

Hardware

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2.1 Hardware

It is the physical computer device which is used for processing data. Hardware includes the CPU, KeyBoard, Monitor, Hard Disk, Printer etc.

2.2 Analog, Digital and Hybrid Computer

There are two main type of computers – Digital and Analog. The third is hybrid computer which is acombination of the above two.

Digital computer operates directly on data which is represented in decimal digit form. This data may benumber or characters which are fed through input devices like punched card, key board or magneticmedia. Digital computers have the ability to handle large volume of data by way of storing them in themachine in the Binary form i.e. 0 or 1, process them and generating reports. Digital computers are used incommercial data processing or work like research mainly in the form of mathematical operations. Example– Any business computer.

Analog computers takes input in the form of electrical pulse. The input are like voltage, pressure, current,water flow etc. The process of feeding data is an analog computer is predefined and operation with thedata and the output follows a standard set of guidelines. Outputs are generally a graph or a picture or atable and give signal for control of operation. These computers are generally used in industrial field incontrolling various physical parameters or stimulating them to find the optimum solution.

Digital Analog

1) Operation on number 1) Operation on physical quantities

2) Process data 2) Measures the physical parameter

3) Generate reports 3) Provide control information

4) Discontinuous 4) Continuous

5) Mainly commercial and scientific work 5) Mainly process control unit.

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Hybrid computer – It is a combination of both digital computer and analog computer. Generally, in thiskind of system data are fed in analog form and output is generally in digital form.

2.3 Evolution of Hardware

As usual, the development of computer has come through evaluation process. The phase of evaluation hasbeen distinguished by quantum jump in technology and they are termed as generations in developmentprocess.

1. First Calculating Machine (450BC) – The Abacus is known to be the fast calculating machine whichwas invented by a Chinese mathematicians named Abacus. It was very primitive in its look andconsisted of a frame and a number of wires.

2. Napier’s Bone (1550-1617) – This was invented by a Scotish mathematician – John Napier. It usedrods which were similar to human bones. This gave the rise to the name Napier’s Bones. It couldmultiply the numbers with the help of rods.

3. Adding Machine (1623-1662) – This was invented by French mathematician Babie Pascal. This machinehad the advantage that it could carry digits and used to add digits. The principle used to add numberswere mechanical counters.

4. Analytical Engine (1791-1871) – Charles Babage, an English mathematician developed this machine,this machine could perform a great deal of calculations. The machine had the following features:-

• Punch card system as input device

• External memory

• Power to do arithmetical calculations

• With a speed one addition in one second

These features are fundamental in nature and in true sense had a great pioneering work in the filed ofdevelopment of computer. For this reason Charles Babage is called the Father of Modern Computers.

5. Punch Card Machine (1889) – Dr. Hollorith, and American developed the punch card machine. Hewas involved in the American census in 1890. The result of census used to be greatly delayed thosedays in absence of sophisticated calculating machine. He devised a card in which holes to be punched.There were wire brasses in the machine. The holes in the card enable the wire to touch a metal platewhich carried an electric charge. The charge was transmitted to electric counters, thus the punchcards could help to identify the figures through the difference in charge. Modern punch card machinewere developed from it. In modern punch card system there are 80 columns and 9 rows. Uniquecombination of holes were thought to represent numbers, alphabets and special characters.

6. Eniac (1944) – This was the first electronic calculator devised by Mark I, scientist of Howard University,USA. It could perform all calculations that a small calculator of today can do with a speed of 300multiplication per second. But the size of the machine was too big. It needed two rooms to accumulateand its weight was approx. 5 tonnes. It used vacuum tubes.

7. Univac (1951) – Sparry Rand Corporation of USA developed first commercial computer to be namedas UNIVAC I. The name was after universal accounting company established by Ackert and Mauchy.

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2.4 Different Generations of Computers ?

Continuous research has resulted a significant development in the evaluation process of computer hardware.The basic objective of the research was to bring improvement in computing power, bring more versatilityand reduction in cost. The development process was marked in distinguished jump in technology andeach stage is termed as generation. There has been vast improvement in computer system in terms oftechnology, speed, storage, reliability, size and cost. The table below will give a picture how the developmentin different generations made a difference in technology and efficiency in computing power.

1st Generation 2nd Generation 3rd Generation 4th Generation

Technology Vacuum Tube Transistors Integrated Circuit Large ScaleIntegrated Circuit

Memory Capacity 10,000 – 20,000 chrs 4,000 – 64,000 chrs 32 KB to 8 MB 1 MB to 128 MB

Internal Operating Mili-seconds Micro-second Nano-secondSpeed

External speed - Few thousands One million Ten millionsinstruction/sec

Mean time Minutes Days Weeksbetween failure

Memory Cathode Ray tube Magnatic Core Magnetic core

Peripherals Punch Cards Punch Cards Punch Cards Punch CardsPaper Tape Paper Tape Paper Tape Paper TapeMagnetic Tape Magnetic Tape Magnetic Tape Magnetic TapeMagnetic Disk Magnetic Disk Magnetic Disk Magnetic DiskMagnetic Drum Magnetic Drum Magnetic Drum Magnetic DrumPrinter Floopy diskette Floppy diskette Floppy diskette

Printer Printer Compact DiskVDU VDU Printer

OCR VDUMICR OCRPlottere MICR

Plottere

Operating System Batch Processing Multiprogramming Multiprogramming Multiprogrammingand time sharing and time sharing and time sharing

Real time Real timeRemote processing

Language Machine Language Assermbly Language FORTRAN II & IV FORTRANHigh Level Language COBOL COBOLFORTRAN & COBOL BASIC BASIC

ALGOL ALGOLPASCALPL/1 ETCADA

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1st Generation 2nd Generation 3rd Generation 4th Generation

Application Area Payroll Payroll Payroll PayrollInventory Inventory InventoryBilling Billing BillingAccounting Accounting Accounting

Market Research etc Market ResearchAir Line Reservationetc

Important UNIVAC – I & II IBM 1401, 1620, 7090 IBM – 360, 370Computers IBM 650, 701 HONEYWELL 400 ICL – 1900, 2900

BURROUGH – E101 BURROUGH – 200 UNIVAC – 1100UNIVAC – 1004 CDC – 6600CDC – 1604 PDP–11, VAX 750

BURROUGH 6700

Size & Cost Large Costly Reduced size Smaller size Very Smaller sizeHigh power Increased reliability Less Costly Less Costlyconsumption Less power Powerful system Powerful system

consumption Less power Less powerconsumption consumption

2.5 How a Computer Functions

To function properly, the computer needs both hardware and software. Hardware means the machine i.eelectronic and mechanical devices and software means programs. We can see the physical presence ofhardware where software is stored in memory or in storage devices. If compared with human, body is thehardware and intelligence is software.

Computer does the following four functions :

• Receive the input data through input devices

• Process the data with the help of CPU

• Generate the output through output devices

• Stores the input/output in storage devices

To understand how it works let us see what are the main components of computer hardware and theirfunctions. The components are broadly divided into CPU, Memory and peripheral devices. Peripheralsmeans input or output devices.

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If an analogy is drawn, it will be easy to understand.

Computer System Human

Computer BodyPeripherals Hands & legsVDS EyesInput/Output devices MouthInput devices EarControl devices BrainMemory Memory

2.5.1 Central Processing Unit (CPU)

The central processing unit is the most important component of a computer system. It is called the heart ofcomputer system.

CPU consists of three components :

(i) Control units

(ii) Arithmetical and Logic unit (ALU)

To understand how these units function, two terms must be known first. They are Registers and Addresses.

Registers: Registers are storage locations in Control Unit and ALU. They are used to hold data andinstructions temporarily. The contents are handled very fast to make processing faster.

Addresses: Each Register is identified by its address. The contents of a register can be referenced in futureuse by its address. The addresses of the registers are stored in main memory in address-registers, simplycalled address.

Data FlowControl Flow

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2.5.2 Control Unit

It is the nerve center of the computer system. Its function is to interpret and execute the instructions.

• It controls, supervises all activities of a computer and monitors the execution of programs.

• It coordinates various parts of computer system like ALU, Main memory and Peripheral devices.

• It controls transfer of data from input devices, storage in memory, movement of data from memory toworking space and back and finally to output devices.

• Stores programs, data, and results in main memory in separate partition made for them.

• Permits the users to wants to have access to computer simultaneously.

In the first phase, Control unit takes an instruction from the primary memory and stores it into instructionregister and interpret the instruction with the help of instruction’s operation code table. The time requiredto perform this is called instruction time.

In the second phase, CU actually performs the task as indicated by the instruction. The time required forthis is called execution time.

The execution of job of CU may involve transferring data from one location to another and/or computationby ALU.

Cycle time = instruction time + execution time.

Cycle speed has gone up with the development of computer. First generation computers used to measureit in terms of milliseconds, the fourth generation in pico-seconds and fifth generation in giga-seconds.Supercomputers execute one billon instructions per second.

2.5.3 Arithmetic and Logic Unit (ALU)

There are three gates in ALU – mathematical gates, logic gates and register.

A gate is an electronic switch. Mathematical gates perform the mathematical calculation, logic gates dothe logical operations and register are used for temporary storage of data.

Logical operations are basically comparing, selecting, matching of data whereas Arithmetic operationmeans addition only. In computer, subtraction is addition of complements, multiplication is repetitiveadditions and division means repetitive subtractions.

2.5.4 Primary Memory or Main Memory

Primary memory is for storing program, data, and doing the processing. In fact, it can be visualized asconsisting of five areas for storing different sets of programs and related data as given below:

(i) Operating System i.e. Supervisor Program

(ii) Program (software) storage area to store program for execution

(iii) Input storage area for holding data from input devices

(iv) Working storage area for processing the data

(v) Output storage area for temporary storage of results as a buffer which are sent to output devices.

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Major elements of a Central Processing Unit

Control Unit ALU Main Memory

Instruction register Addition Operating System

Operation Decoder Subtraction Application software

Address Register Multiplication Input data storage

Division Working Storage

Comparing Output storage

2.6 System of Storage in Memory

In Computer Fundamental unit of storage is binary digit or bit. Binary means two. Digital computer usedmagnetic core for memory. These magnetic core are iron-ferrite doughnuts. Binary numbers is representedby a combination of 0 and 1. In Computer memory, “0” and “1” are represented by off and on in theelectrical circuit. In digital computers, memory location consists of electric circuits.

A Computer can store data or instructions through electronic switches in the sequences of off and onpositions in the memory. All digits, instructions and commands are converted into a sequence of binarydigits or bits (0 or 1). Most of the computers today represents numbers, letters or special characters bycombination of group of bits.

The most common group of combination is eight-bit group which is called a byte.

The memory capacity of computers is measured by kilobytes (KB), Megabytes (MB), Gigabytes (GB).

1 kilobyte = 1024 bytes = 1000 bytes (approx)

1 megabyte = 1024 kilobytes = 1million bytes (approx)

A single byte can store one character only. Thus, number of bytes of computer memory required to storea data item is equal to the number of characters of the data field. For example, ten character name fieldrequires 10 bytes.

However, some of the computers have standardized the number of bytes assigned to a specific type ofdata. These standards may be different for different manufacturers. Basically, the combination of bytes areused for storing a large number to increase the accuracy (precision). In case of data in scientific researchneeds higher degree of precision than what is required in commercial application. A computermanufactured for use in scientific research will follow bigger byte group for storage of data and processing.

No. of bytes taken together form a word.

Commonly used word length is 8, 16, 32,64 bits. During processing computer uses bits of its word lengthsize. This is why word length is taken as a measure of computing power. The higher the word size, themore is the power of the computer. A 32-bit computer indicates its word size.

Magnetic Core Memory – It is a small ring of ferromagnetic materials which can be magnetized very fastand retain it until changed. It is used to store binary bit 0 or 1.

Semiconductor memory – Semi-conductor memory consists of electronic circuits. It is made in silicon chips.Today all modern computers use semiconductor memory for the following advantages:

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• Physical size is small

• High capacity

• Low Access time

• Low cost

• Low power consumption etc.

2.7 Different Types of Memory

2.7.1 RAM (Random Access Memory) : The term RAM itself indicate that this kind of memory allows toreading and writing both possible random access this making it very fast. RAM is volatile in the sense thatthis type of space is used for temporary working area and contents are lost as soon as the power in themachine is switched off.

There are two types of RAM :

- Dynamic RAM (DRAM)

- Static RAM (SRAM)

The difference in DRAM and SRAM is in technology in terms of holding data. DRAM has to be refreshedfrequently (thousands in one sec) whereas refreshing need of SRAM is very less.

2.7.2 ROM (Read Only Memory) : The term itself indicates that this type of memory is used for permanentdata storage and allows only reading. ROM’s are normally used to store information which is used by thecomputer for its own operation. Unlike RAM, ROM is not volatile i.e. contents of ROM is not lost as soonas switch of the machine is made off. Computers uses ROM chip to store permanent program for exampleprogram for booting the computer. At the time of manufacturing of ROM chip, program is stored.

Difference between ROM and RAM

ROM RAM

Allows only read facility Both facilities – Read and Write

Contents are permanent Volatile – i.e. lost when power is switched off

Contents are fixed which are It is an working space for the computer.generally for use by computerfor its own operation

2.7.3 PROM (Programmable Read Only Memory): PROM chip is used to store program of permanentnature. PROM is used for microprocessor dedicated for a particular job. At the time of manufacturing,PROM chip remains blank and program is written on it afterwards.

2.7.4 EPROM (Erasable Programmable Read Only Memory): This kind of ROM is used for storing programs.It is different of PROM in the sense that earlier program can be erased by lesser rays and new program canbe stored there.

2.7.5 EEPROM (Electrically Erasable Programmable Read Only Memory): This is a special type of PROMwhich can be erased by electrical charge.

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2.7.6 Flash Memory : It is a special type of EEPROM where programming is faster (in blocks). Flash memorychip is used only to keep provision of faster updation.

2.7.7 Cache Memory : This is a very fast RAM. It is costlier than the RAM cost in terms of capacity. Thebasic objective of using it is to do processing faster. It is placed between CPU and main memory. Whensome information is needed frequently for processing, the same is copied from RAM to Cache memory tohave faster access to the contents of cache memory and thus enhancing the speed of processing.

2.8 Secondary Storage Device

The main memory of a computer is limited in size and expensive also. Thus, main memory is never meantfor the entire data storage. The data or programs whatever required during processing are loaded into thememory and after processing they are stored back in the storage media for subsequent use. This provisionfor additional storage space is called secondary storage.

It is like the provision made by the human being. A person can not keep everything in his memory.Something is stored in his note books, files, books etc. He consults them whenever required. Exactly in thesame fashion, secondary storage is a space or media in which data or programs are stored for use at thetime of need.

The most common secondary storage device are floppy disk, magnetic disk, magnetic tape etc.

Characteristics of secondary storage device :

1. Mass storage – Large volume of data can be stored. Now in a hard disk, 1 GB to 8GB data can bestored. In a magnetic tape, 150 MB to 250 MB storage capacity is very common.

2. Back up provision – This is a provision of storing of important data files or software in storage devicefor protection from damage.

3. Non-volatile – Data is not lost due to power failure.

4. Movable storage media – Storage of data in secondary storage device gives a very important facilitiesthat the data set can be ported from one place to another, one machine to another.

2.9 Virtual Memory

Virtual memory is a provision of secondary storage which acts as primary memory. When the size ofmain memory is less than required size for a program to run, the operating system enables it with the helpof hard disk (main storage media). The system followed is breaking the programs and accommodatingthe part according to the size of main memory available and rest is stored in hard disk. When a part ofprogram stored in hard disk is required for running the program, a part of the program stored in the mainmemory is transferred to hard disk and the required part is brought to main memory. This process iscalled swapping.

2.10 Classification of Digital Computers

In terms of capacity, computers can be classified as Mainframe, Mini and Micro Computers.

Mainframe Computer – It is a very big computer with computing power. In a networking environment,mainframe computer is the nodal point of network.

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Mini Computer – It is a smaller version of mainframe computer both in size and computing power.

Micro Computer – Micro computers are built with the help of micro-processor chips. CPU, ROM, RAM arein different chips. Earlier days, micro-computers used to have limited power. Today, power of micro-computers have increased enormously.

Main frame Mini Micro

Computing power Maximum computing Moderate computing Built with micro processorpower power chips

Size Big size Smaller in size Very small in size

Operating System Operating System Operating System Single user operating systemsupporting supportingMultiprocessing or Multiprocessing ormultiprogramming multiprogramming

No of peripherals Large number Less no of peripherals Minimum No of peripherals

Cost High Less Cost Low cost

Application Network Business Applications Business ApplicationsData warehousing Government Department Government Department

Education etc Education etcStatelites in network

Example IBM 5/390 VAX 8200 of DEC Personal Computer orWorkstation

Further, there is another kind of computer which has fastest computing power and huge storage devices –called supercomputer. The speed of this kind of computer is measured in terms on nanoseconds and evenpico-seconds which can perform 2 billion mathematical calculations per seconds. Example – Y- MP/C90of Cray Research Inc.

2.10.1 Personal Computer (PC)

Personal computer refers to stand alone mode computer which is fully equipped with own CPU, memory,storage device, operating system and utility software etc. It is based on microprocessor based technology.The use of PCs have revolutionized the computer world today because of its versatility. It can used aspurely a home computer or an intelligent terminal or a stand alone system capable of undertaking vastload of processing which has a capacity of a mini computer.

2.10.2 Workstation

Workstation generally refers to an intelligent terminal in a networking environment. Workstations may beany computer linked under LAN also. Sometimes, workstation may be single-user computer with highstorage device. The way workstations are being termed these days, it can be taken any computer meant fora unit taking care data entry or data storage load in a system to be a part of a bigger processing infrastructure.

2.10.3 Laptop Computer

A laptop computer is small, lightweight and portable. It has all the features of a PC. Because of compactnessin the architecture and limited production, the cost is still high.

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2.10.4 Notebook Computer

A Notebook Computer is typically of size of a note book and can be fit in a brief case. It has CPU, Memory,disk drive of same capacity of a PC and a display screen. It is manufactured based on flat panel technology.

2.10.5 Hand-held Computer

It a small computer which can be held in hand. Hand-held computers are also called as palmtops, pocketcomputers and PDA (Personal Digital Assistant).

2.11 Input / Output Devices Input Devices

Input devices are the machine used for feeding data into a computer. Example: Keyboard, floppy disk,terminal etc

Output Device: Output devices are the machine through which output of processing comes out from thecomputer. Example : Monitor, printer, floppy disk etc.

There are some devices which can be exclusively used as input device like Keyboard, Card Reader, CD-ROM etc.

Similarly, there are some devices which can be exclusively used as output devices like Monitor, printer etc.

There are devices which can be used as both input and output device. Example: Floppy Disk, MagneticTape, Magnetic Disk etc.

2.12 Different Types of Input Devices

2.12.1 Key- Entry Device: It is a device like a type-writer called Key board for data entry. The standard keyboard is qwerty Key Board. Most of the key-entry system is off-line device i.e. not connected to the centralcomputer.

2.12.2 Punch Card System: This was very popular data entry system in 70’s. This system became obsoletenow. Hollerith cards were used in electric Punch Machine and the coding used there is called Hollerithcode. Punching means making holes on the punch card. Punch card is of 80 columns. The combination ofholes in different columns used to indicate different numbers, characters and special characters.

2.12.3 Floppy Diskette System: The floppy diskette is used for data entry and data storage. This system ofdata entry device is very popular today. Data is entered through the keyboard, video screen displays theinformation. The advantage of the system are :

• Input data can be edited in the floppy.

• Floppy is reusable.

• Validation of input data can be done at the data entry through programming.

• Messages on the screen are helpful for operators.

• It is electronic device whereas the punch card system is mechanical one and thus more reliable.

• The system is faster and gives high productivity.

2.12.4 Terminal: This system has become very popular today. Most of the computers have terminals. Theseterminals are used for data entry. Every terminal has a visual display unit with it. The input data are

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displayed on the terminals. Now a days, data entry is done with the help of user-friendly program. Theprogram displays menu and options can be selected according to the choice of the operator. Data is validatedon-line and with the help of master file, typing load can be minimized.

Advantages :

• Relatively inexpensive.

• During entry, data is visual.

• Data goes directly to the computer i.e. no intermediate storage.

• Have advantage of operation from widely dispersed geographical areas.

Disadvantages:

• If the computer goes down, terminals becomes useless.

2.12.5 Dumb Terminals and Intelligent Terminals: A dumb terminal is used for data entry only. It has thefacility to communicate data to and from CPU. An intelligent terminal, on the other hand, has the followingfacilities :

• It has a CPU to control the function.

• It has buffer memory.

• It has the ability to attach the peripherals i.e. floppy disk, printer etc.

2.12.6 Point of sale (POS) Terminal: These are very small terminals like cash register. Under POS datacollection system, sales data are entered at retailer place. Retailers requirements are entered and the sameinformation reaches to the warehouse immediately so that distribution of the product can be made faster.

Advantage of POS data collection :

• Facilitate faster transaction.

• Standard procedure of calculation of sales price, tax etc. can be performed readily and the same can bemade to the retailer.

• Update of inventory position can be known to the salesman instantly.

2.12.7 Mouse: Mouse is not in true sense a input device. Mouse is kept on a rubber pad called mouse pad.Mouse has a rubber ball at the bottom and two or three switches. Movement of cursor can be controlledwith the help of mouse and switches are used for different options. The operator can operate faster byreaching a particular point on the screen and pressing the button. Pressing button is called clicking. Now-a days, the user friendly software or even operating system gives the menu on the screen and operator canactivate the required program by choosing the particular option of the menu or a box by simply bringingthe cursor to the appropriate point on the screen with the help of mouse and clicking it.

2.12.8 Punched Paper Tape: This system consists of long roll of one inch wide paper. A vertical column ofthe tape is used to represent a single character. It is rarely used.

Problems in the paper tape:

• Paper tape is not reusable.

• Its capacity is small.

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• Paper tape is not convenient for use.

• It is slow in transferring data.

2.12.9 Track Ball: Its function is same as mouse. The difference is that, it is not required to move to controlthe movement of cursor. What is required is to rotate a ball on the device with the help of thumb to controlthe movement of cursor.

2.12.10 Joystick: This is device is similar to mouse. The difference is the cursor continues to move in thedirection joystick points. To stop the pointer, joystick has to brought back in upright position.

2.12.11 Light pen: Light pen consists of photocells which activates some desired action when the usertouches the PC screen with it.

2.12.12 Magnetic Ink Character Recognition ( MICR): MICR code is developed by American BankingAssociation ( ABA). The advantage of MICR code is that any document having MICR character can beread by a computer because both manual entry data and data through reading devices are checked toensure correctness. Another advantage of MICR is that it reduces or eliminates the amount of codingunnecessarily.

2.12.13 Optical Character Recognition (OCR) Device : It is a reading device used to interpret printed,hand-written data directly from source documents. OCR device make use of light-sensing mechanism andlaser technology to interpret recorded data. The essence of OCR is pattern recognition. For this reason, thesimpler the character front, the more reliable the OCR system. OCR capability of an OCR device is veryhigh i.e. in the range of 2000 documents per minutes. Machine is costly. It is cost-effective only where10,000 or more documents have to be read per day.

2.12.14 Scanner : Scanner is a very useful device for reading text or documents. The scanner digitizes theimage of the document. Now a days, the optical scanner system supports a software – OCR package (OpticalCharacter Recognition Software). The software translate the image into ASCII characters.

Types of Scanner :

i) Optical Scanner – It reads text or data with the help of light source and light sensors. Optical scannersare Optical Character Reader ( OCR) and Optical Mark Reader (OMR).

ii) Magnetic Ink Character Recognition (MICR) Scanner – It reads character imprinted on a documentusing magnetic ink containing iron oxide. It is generally used in Banking Industry for clearing ofcheques.

2.13 Different types of Output Devices

2.13.1 Monitor : Monitor is used to display the data or commands entered through Keyboard, anymessage, graphical picture, table etc from the machine. It is synonymously called VDU (Visual DisplayUnit). A digital monitor uses CRT technology. A digital monitor translates the digital signals intoanalog signals which controls the display. Different adapters are used to have control on video standard.They are:

• MDA (Monochrome Display Adapter)

• CGA (Colour Graphics Adapter)

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• EGA (Enhanced Graphics Adapter)

• VGA (Video Graphics Adapter)

• SVGA (Super VGA).

In earlier days, the display unit on the console too small. With the advancement of the technology, thefeatures of the monitors have been developed and it is a very effective unit for communication betweenman and machine. Rather, advancement and the success of user friendly system could be imagined onlyafter development of monitors. Depending on the ability of displaying matters in colour in the monitor isclassified as given below:

Monochrome – One colour for backgound and one colour for foreground. Common colour is black andwhite. But it can be green and white or amber and white.

Colour Monitor – Colour monitor can display 16 to 1 million different colours.

For example a 8-bit monitor uses 8-bit for each pixel. 8-bit pixel can display possible 256 colours. ( 256 = 2to the power 8). More bits per pixel, more will be range of colour combinations. A pixel represents a singlepoint.

Size of Monitor: Common size is 14 inch. Bigger than this is also available.

Resolution: Resolution indicates the density of the pixels. Quality of display depends on resolution.

Popular VDU for PCs

VDU Resolution Simultaneous Colour

VGA 640 x 480 16

SVGA 800 x 600 16

1024 x 768 256

1280 x 1024 256

1600 x 1200 256

8514/A 1024 x 768 256

XGA 640 x 480 256

1024 x 768 256

TI 34010 1024 x 768 256

2.13.2 Printer:

The printer is the device that prints the output of the computer on paper.

Different types of printers available as follows:

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Type of Printer Features Speed Quality of Print

Line Printer The characters of a line is 300 – 1200 lpm Ordinaryassimilated and printing of onefull line is done at a time.Can only print text.

Dot Matrix Each character is created by striking 120 – 600 cps GoodPrinter pins on the printer head. Printing is

done character by character.Can only print text.

Daisy Wheel It uses a disk ( plastic/metal) on 10 –75 cps GoodPrinter which characters are set. The printer

rotates the disk to bring desiredcharacter to print. Printing is donecharacter by character.Can only print text.

Ink Jet Printer It spays ionized ink at a sheet of 300 dot per inch Very Goodpaper. Magnetized plates gives theshape of ink path.Can print text in different fronts andgraphics also.

Laser Printer It user a laser beam to produce image 1200 – 2400 dpi Very Goodon a drum which rolls through atoner to print the image on the paper.Can print text in different frontsand graphics also.

Impact or Non-impact Printer

Impact printer is the printer which prints by striking the ink ribbon. For example – Dot line printer, Matrixprinter, Daisy wheel printer.

Non-impact printer are Ink jet printer, Laser printer, Plotter.

2.13.3 Magnetic Tape

Magnetic Tape used in computer is like the one is used our home tape. Information records in a magnetictape is kept by a combination of magnetic spots arranged in a number of horizontal tracks. Size of thestandard reel is 2400 ft long. But tape of length 1200 ft and 3600 ft are also available. Reading or writing ofmagnetic tape is done by magnetic tape drive.

Different type of tape drives are:

• Vacuum column drive

• Steamer tape drive

• Cartridge tape drive

• DAT cartridge

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Type of Tape Storage Capacity Uses

½ inch steamer tape 40 MB – 100 MB These are used in vacuum column or steamer tapedrive. Drives are costly but the tapes are no costly.

½ inch cartridge tape 60 MB – 600 MB Tapes are cheap and used in cartridge tape drive.

¼ inch 40 MB to 5 GB Tapes are cheaper and used in ¼ inch cartridgetape drive. Data transfer rate is high.

8-mm tapes 1 GB – 5 GB They are used is 8-mm helical-scan cartridge tapedrive. Data transfer rate is slow.

4-mm DAT 2 GB – 24 GB They are used in DAT cartridge tape drive. Datatransfer rate is slow.

Magnetic tape can be coded in different formats. i) Seven track ii) Nine Track.Nine track is preferable for the following reasons:

i) 8-bit character code such as ASCII or EBCDIC and parity bits provides better accuracy in reading orwriting of tape.

ii) It can store pure numerical data in packed decimal format i.e two numeric digits are in one column ofa tape.

Density of magnetic tape is measured in terms of bits per inch (bpi) or character per inch (cpi). Commontape densities are 800 bpi, 1600 bpi or 6250 cpi.

To separate one record from another record, inter record gap (IRG) is maintained in a tape.In fact, interrecord gap occupies a good amount of space. To reduce the wastage of space, records are created in blockswhereby some records taken together are recorded as block. This saves lot of space.

There are two marks on the tape. Beginning of tape (BOT) and end of tape (EOT). Header and trailer levelsare maintained for verifying the accuracy and compactness of file processing. A file protection plastic ringis available is called read/write ring. Removing the ring means in read only mode i.e. writing is notpermissible.

Advantage of Tape :

• Cost effective

• Huge storage capacity

• Easily stored in data library

• Durable back-up media

• Reusable medium

• Fast input/output

Disadvantages of Tape:

• Reading a specific record is to be done by a time consuming sequential process.

• Data transmission is slow in comparison to disk.

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• File protection is lost if the write-protection ring is with the tape.

• Variations in tape densities causes serious problem in portability. Tapes recorded at a very low orvery high densities sometimes are not readable by other tape drives.

• File updation (record insertion or deletion) is not possible on the same tape.

• Tape rewinding is a must for dismounting.

• Difficult to recover the files from parity error or physical damage of tape.

• Data can be accidentally erased or overwritten.

2.13.4 Hard Disk

Magnetic Disk consists of circular metallic plate coated with magnetic oxide. It resembles with gramophonerecord which can be mounted on a spindle. Recording is done on both sides of the plate. The set of diskplate and spindle is called disk pack.

The term hard is used to distinguish it from soft disk i.e. floppy disk.

Disk Pack is mounted on a disk drive for reading from writing on it with the help of a read - write headattached with the disk drive. The read-write heads are attached with a common access arm which movesthe read-write heads simultaneously to the position of the disk where reading or writing takes place.

The read-write heads of a magnetic disk do not touch the recording surface. There is a very small gapbetween the read-write head and the recording surface which is called head’s fly height. If the read-writeaccidentally touches the surface or the gap of head’s fly height gets jammed by any microscopic particle,the entire disk pack gets damaged and we call it as head crash.

Types of Disk Drives :

i) Exchangeable – This type of disk are replaceable. Once the work on a particular disk is over, it can beremoved and another disk can be mounted.

ii) Fixed Disk – This type of drive is permanently mounted in the computer. This is also called Winchesterdisk. The term Winchester came from the place Winchester where this type of drive was first developed.

How recording is done in a magnetic disk

On the surface of the disk, there are concentric circles called tracks. Data characters are coded into bits andwritten on the tracks. Like magnetic tape, magnetic disk also use parity bits for accuracy of data. Beginningbit and ending bit of a character are marked but there is also inter-record gap as in the case of magnetictape. One track is divided into number of sectors having same capacity. In a disk pack, let us assume

Recoding surface = 10

No. of track in each surface = 200

No. of sector in each track = 19

Then, total No of sectors = 10 x 200 x 19 = 38,000

In a disk, each record has a unique address in terms of sector. With the help of unique address, itbecomes easy to fetch a particular record. This is why, disk is called Direct Access Sector Device(DASD).

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2.13.5 Floppy Disk

Floppy disk is also soft disk. It is smaller version hard disk and has advantage of handling and portability.It has two different sizes – 5 ¼ inch and 3 ½ inch.

Type of floppy Storage capacity Description

5 ¼ inch 360 KB or 1.2 KB Common density in use – double density and high density.

3 ½ inch 720 KB or 1.44 MB Size is small but capacity is higher.

2.13.6 ZIP Disk

These are high capacity floppy disk like device. It is little larger than 3 ½ inch floppy disk and thickness isalmost double of floppy disk. Their capacity is 100 MB. Because of its small size and high capacity, itspopularity is on rise. The technology of laser beam has made it possible to write data in tiny pits. To readthe data, laser beam is used to scan the same. This laser technology has evolved optical disks for use incomputer. Two popular types of optical disks are CD-ROM disk and WORM disk. The advancement oflaser technology may bring revolution in data management in Computer system in future.

2.13.7 CD-ROM Disk

Its full name is Compact Disk – Read Only Memory. The name itself explains that the data are stored canbe read only. For reading the data fron CD-ROM, a laser scanner is required. CD-ROM drive supports thesame and for that reason it is costly. It is popular because of its high capacity which is 630 KB – 1 GB.

2.13.8 WORM Disk

The term stands for write-once, read many. These are metal film disk on which data are written with thehelp of laser. In fact, the difference between CD-ROM and WORM is that WORM allows to write once forthe first time. Subsequently, the behaviour of both is same.

2.14 Selection of a Computer System

2.14.1 Criteria of Selection of a Computer:

Following criteria should be considered for selection of a computer:

i) Workload – data processing requirements in terms of volume, frequency etc.

ii) Operating system capability.

iii) Type of application system to be processed.

iv) Processing requirements – batch or on-line or distributed data processing etc.

v) Network requirement.

vi) File/data base design support – their special features.

vii) Software performance, capability and special features.

viii) Required data reliability, security and integrity.

ix) Reputation of manufacturer.

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x) Reliability of machine.

xi) Cost of machine, delivery, installation.

xii) Maintenance cost etc.

2.14.2 Procedure of Selection of Computer

The selection of computer should follow a systematic procedure to assess the performance of computer,matching the requirement specifications of user, reliability and cost. The standard procedure is:

i) Preparation of design specification.

ii) Collection of offer from vendors.

iii) Conducting benchmark test to understand performance.

iv) Preparation of comparative statement with the following:

v) Benchmark test results

• Configuration

• System software features

• Cost

• Maintenance provisions

• Other terms

vi) Negotiations with vendors, if required

vii) Decision on selection of vendor and fixation of schedule of installation

2.15 Hardware Procurement

Hardware may be procured with the following options:

1. Outright Purchase

2. On Rent

3. On Lease

4. Hire Purchase

Outright Purchase option are generally practiced in India. The advantage in outright purchase is flexibilityin the use of the machine according to the need of the organization. Moreover, tax benefit on depreciationis another point which encourages the users to go for outright purchase.

Hardware is procured on rent for the following advantages:

i) Cash outlay for procurement is avoided.

ii) Rent is allowable expenses for tax.

iii) Risk of obsolescence is not with the user.

Procurement on Lease is opted in case of machine is of high configuration and seller is

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generally the manufacturer.

i) Cash outlay for procurement is avoided.

ii) Lease charges is allowable expenses for tax.

iii) Risk of obsolescence is with the lessor.

Hire Purchase option is normally not used as Computer becomes outdated very fast. This option isexercised only to avoid immediate investment for procurement.

2.16 Installation of a Computer System

The following arrangement/facilities must be made for installation of a computer:

• Site preparation with sufficient accommodation for machine and peripherals with flexible provisions

• Good height of the room.

• Proper electrification.

• Fire protection arrangements.

• Air conditioning.

• Provision for controlling temperature & humidity.

• Protection from dust .

• Proper environment control – free from air pollution, noise, excessive vibration, electromagnetic effectsetc.

2.17 Hardware Maintenance

Hardware maintenance means maintenance of computers, peripheral devices and network facilities.Maintenance may be Preventive Maintenance or Breakdown Maintenance or Replacement Maintenance.Maintenance cost is also a consideration. In case of hardware maintenance, proper care is taken to preservethe data in the computer.

Preventive maintenance means provision of mandatory maintenance with certain periodic interval as aprecautionary measure to reduce the risk of failure or breakdown of computer. The interval may be aweek or a fortnight or a month. The interval is selected depending on the recommendation of manufacturer,sensitiveness of failure etc.

Breakdown maintenance means maintenance after actual breakdown of the computer. The failure of aparticular module of the computer or the peripherals are diagnosed in a systematic manner and generallycards or chips or parts are replaced to put back the machine in working condition.

Replacement Maintenance involves replacing major devices of the computer. This happens when thatdevice becomes bad or the same needs up-gradation. In case of replacement, proper back-up of old datafiles and restoration is a important consideration.

Software andit’s Tools

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Software & its Tools

3.1 Software

Software means program. Generally, the term software indicates a software package that is a set of programswhich in combination does a particular job. Without software, the computer hardware cannot function.Software actually activates the hardware to work.

3.2 Computer Program

A computer Program is a set of instructions given to the computer to perform a job in orderly fashion. Aprogram is developed in a particular language. The language has a set of command words and specialgrammar for construction of commands. Writing a program means writing the commands in the sequenceof flow of logic of the program.

There are basically two types of software – System Software and Application Software.

3.2.1 System software is the type of software which controls the function of computer system. It may agroup of software used for managing the different functions of Computer system e.g. allocation of resourcesto different jobs, control of input/output, device management, memory management, translation ofprogramming language to machine language, command interpretation etc.

The categories of software which are termed as system software are :

• Operating System

• Utility programs

• Programming Languages etc.

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3.2.2 Operating System : It is a collection of programs designed to maximize the working capability of acomputer, by way of scheduling and supervising the execution of programs, allocating storage and input/output devices and handling errors etc. Operating software may be defined as a set of programs developedto control the execution of other programs like application programs or utility programs.

Example of some popular Operating System :

Microsystem - Windows, DOS, OS/2, Lunix etc.

Minicomputer – Unix, OS/400

Mainframe - MVS

Following are the functions of the operating system :

• Job Management – The function includes scheduling of job, activation /deactivation of processesinvolved in the job. This is done on the basis of algorithm of scheduling and availability of processorand main memory for carrying out the job.

• Memory Management – This involves allocation of main memory space to different jobs and keepingtrack of it.

• Device Management – It involves tracking the status of devices, allocation of various devices to jobs,activate them when needed.

• File Management – It involves efficient allocation of separate space for each file, arrangement ofprotection from loss of data etc.

• Interaction with operators – Interpretation of commands from operators, display of error / interruptionmassage etc

• Security – Function of protecting unauthorized access to system. For this purpose security mechanismthrough password is followed.

• Job Accounting – Keeping track of time and resources used by different jobs etc.

Classification of Operating Systems

Operating System may be classified as :

• Multi-user – It allows more than one users at a time. Example MVS, Unix etc.

• Multi-processing – This kind of OS has the ability to process more than one process at a time byallocating multiple CPU’s to different processes. The competitive resources are allocated to differentprocesses depending on their availability. If the resource is not available, it is kept on the hold status.

• Multi-tasking – This kind of OS allows processing of more than one program at a time. The only CPUin the system is allocated to a task when others are allocated peripheral devices to do Input/Outputjob. Example Unix, Windows NT etc.

• Multithreading – This kind of OS allows different parts of one program to run concurrently.

Bootstrapping

Bootstrapping means loading Operating System in Computer after the power is switching on. Bootstrapping

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instructions are stored in RAM. Once the power is on, the computer takes the instruction of bootstrappingfrom ROM and loads the Operating System from Hard-disk into RAM and put the computer in opetration.

3.2.3 Utility Software: These are basically to do some routine jobs of specialized in nature which enhancesthe performance of the computer and /or serve a great utility function to the users.

Examples of Utility software :

• Sort/Merge software

• Editing software

• On line debugging

• Report generator

• Query language etc.

3.2.4 Programming Language: Programming Language is the coded language understandable by theComputer. How a Computer understand the programming language. A Programming Language isdeveloped with its own grammar and features of mathematical tools to handle data. The efficiency of aprogramming language depends on its versatility of use in handling different problem situations, datahandling capacity, file handling features etc. Today high level language or fourth generation language isthe result of extensive research in developing programming to make programming more efficient and itsuse universal. Let us understand the track of development of programming languages.A Computer understand only 0 and 1. In the early days programs used to written using 0 and 1. Job ofmachine was very simple to understand the program. But the skill requirement of programmer was higherand writing program was very difficult job.

3.2.4.1 Machine Language : Programming language was of some codes with 0 and 1 only. That was only

Machine Language Structure :

Operand Code Operand In case of arithmetic operation, the structure was

Operand Code Operand 1 Operand 2

way to make machine understand the program. The job of a programmer was difficult and tedious. Ofcourse, codes were limited in numbers and language had a rigid structure. The biggest difficulty was inreferencing the operands (variables) by their address locations.The operand code was a binary code and operands were memory locations.

Machine Language was machine dependent.

Then the thought process of using some standard code for mathematical operations came and assemblylanguage was developed.

3.2.4.2 Assembly Language : It is called low level language. Structurally it is similar to machine languagebut with some development. Operand code were symbolic and location for operand could be given asymbolic name. For this reason it used to called as symbolic language. It was also machine dependent.

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Operand codes were like

ADD for addition

SUB for Subtraction

MUL for multiplication etc.

3.2.4.3 Assembler : A computer can execute only statements written in machine language. Thus programswritten in machine language is easily executable. But programs written in Assembly language or HighLevel Language needs conversion into machine language. The Conversion job of programs written inAssembly Language is done by a software called Assembler. Assembler does direct translation of Assemblyprogram to Machine Language Program.

Advantages of Assembly Language :

• Operand codes are in symbol. Thus it is easy to remember.

• Programmer need not to keep track of memory locations.

• Fewer instructions in the language with simple structure – easy to learn.

• Debugging is simpler.

Disadvantages of Assembly Language :

• Program is very long

• Processing is slow

• It lacks portability

3.2.4.4 High Level Language ( HLL) : High Level Languages were developed in such a fashion that theinstructions are alike to English Language so that following the instruction codes and writing programsbecome easier. This helped the programmers to overcome the shortcomings of low level languages to agreat extent. The following are the features of High Level Languages :

• They are English like Languages and easy to learn.

• Standard sets of words and well defined structures are used.

• Program development effort is less.

• Debugging is easy.

• Portability is high.

• It is not machine dependent.

• It was procedureoriented thereby reducing programming effort.

3.3 Compiler

Programs written in HLL is called Source Program. Corresponding to each HLL, there is a software calledcompiler. The task of compiler is to convert the Source Program into Machine Language Program to makeit executable. It does it in two stages.

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In the first stage, it checks the syntax errors in the program and prints all syntax errors which are calledcompilation errors with error codes. These errors are corrected by the programmer. Once the program isfree from errors, an object program is generated.

In the second stage, from the object program machine language program is generated. The process ofconversion from Source Program to Object Program is called compilation.

Compiler translates the whole program in one go. Unless the whole program is free from errors, conversionfrom source program to object program is not possible. Once the executable program is generated from thesource program, the same is used for execution. In fact, the major difficulties of low level languageprogrammer have taken over by the compiler.

3.4 Interpreter

It is a type of translator from source program to machine program. Interpreter translates the source programduring execution line by line. When a program is executed under an interpreter, the interpreter checks thecorrectness of a line in the program and if a line is found correct, it translates the line for execution. If theline is found incorrect, the interpreter displays the line and type of error and stops execution. This processslows down the execution process. For each run of the program, the interpreter has to repeat the process ofinterpretation.

Even if a program is successful in running once, it does not ensure complete correctness of the program asfar as syntax errors are concerned. The reason is that during the earlier run, the program might not havenot encountered a condition and have not executed some lines or a subroutine where there may beundetected errors. In the subsequent runs, by encountering some other logical condition the programmight have to execute some other lines or some other subroutine in the program and errors may be detectedduring that point of time.

The cost of interpreter is much less than the compiler.

3.5 Difference between Compiler and Interpreter

Interpreter Compiler

Translates line by line. Translates the whole program.

Translation takes place during execution. Translation of whole program in one go.

Execution time is high. Execution is fast.

Requires less memory. Requires more memory.

Successful run for sometimes does not Compilation is very time consuming process.guarantee full correctness of the program

Cost of interpreter is less. Cost of compiler is high.

3.6.1 Problem - Oriented Language – The focus of this type of programming languages is solving problem.Programs developed in this type language takes data and prints output. RPG is a typical Problem-OrientedLanguage.

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3.6.2 Procedure-Oriented Language – COBOL is a procedure oriented language. A program can be brokeninto modules and for each module a procedure can be written.

3.6.2.1 COBOL – Common Business Oriented Language

COBOL was developed by a committee of conference on Data System Language (CODASYL) in 1959-60.Latest version is COBOL–80. It has been widely used all over the world. COBOL has the following features:

• It has four divisions

� IDENTIFICATION DIVISION

� ENVIRONMENT DIVISION

� DATA DIVISION

� PROCEDURE DIVISION

• The different sections under different divisions have special function.

• The instructions in Procedure Division are almost similar to English Language.

• Provision of separate procedure for separate sub-module is a good feature of modular programming.

Advantages of COBOL:

• Easy to learn

• Very close to English Language adding self documentation features

• Most suitable for commercial applications

• Supports features of Structured programming

• Good capability of file handling

• Procedure-oriented language helping modular programming

Disadvantages of COBOL:

• Program is lengthy

• Not suitable for scientific or mathematical programming

3.6.2.2 BASIC – Beginners’ All-Purpose Symbolic Instruction Code

It was developed by John G Kenedy and Thomas Kurtz, Professor of Mathematics at Dartmouth. This isvery simple language and it is suitable for both scientific and commercial applications. A statement iswritten in a line and line number is assigned to each statement. Flow of control is managed through linenumber.

Advantages :

• Easy to learn

• It is supported by both Interpreter and Compiler.

• Has a good debugging facilities by throwing syntax errors at the time writing program

• Statements are very compact

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• Simpler read/write statement

• Arithmetical statements are simple algebraic equations.

• FOR – NEXT loop and simple sub-routine structure has made the program very versatile.

Disadvantages of BASIC:

• Has limited capability file handling

• Does not support ISAM file

• Not suitable for Structured Programming

3.6.2.3 Fortran - Formula Translation

The language was developed by IBM (International Business Machine) in 1957. There has been continuousdevelopment resulting FORTRAN II and FORTRAN IV, FORTRAN 77. Main features are :

• Statement number is optional. This is adopted to control the flow of program.

• Easy handling of arithmetic operations.

• Data definition provision fixes the size of variables ensuring more precision in results.

Advantages of FORTRAN :

• Easy to learn

• Efficient in mathematical problem solution

• Generally, machine dependent

Disadvantage of FORTRAN :

• Has limited capability of file handling

• Not suitable for commercial application

• Does not support ISAM files

• Not suitable for structured programming

3.6.2.4 PASCAL :

PASCAL was developed by Nikolas Wirth and named after the Great French inventor of Computer, BlaisePascal.

• It has the structured programming

• It does not support the statement GOTO thus allows GOTO-less programming

• Has good ability of both file handling and mathematical calculations.

• Useful for all types of problem solutions.

3.6.2.5 ALGOL

ALGOL ( Algorithmic Language) is basically developed for mathematical problem solving. This languagesupports features of structured language. The Language also supports features of dynamic allocation ofstorage.

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3.7 Fourth Generation Language ( 4 GL)

The technology of software development has undergone revolutionary changes. With the evolution ofsophisticated hardware, the usage of computer has expanded far and wide. To make usage of computermore popular, the tools of software development has been made simpler. Development of wide varietiesof data structure is a part and parcel of the evolution process. To reduce the burden of programmer,programming language evolved some user-friendly tools. Like the evolution in hardware, programminglanguage is undergoing developments which has conceptualized as generation in terms of features,capability and technique. The Fourth generation language are bracketed as the languages which have thefollowing features:

• It is a combination of simple commands and simple structure.

• It includes decision support facilities with simple what-if structure.

• It supports tools to develop user-friendly software

• It has facilities of easy handling of databases

3.8 Some General Software Features

Now a days, software are being development with broad based features for general use and thus theybecome complex. Apart from it, keeping in view the size of the application and the cost , an application likeERP may have four versions – desktop, standard, professional, enterprise etc. The software are to be user-friendly, network friendly and so on. To add these features is not simple. Some common features of today’ssoftware are given below :

1. Graphical User Interface ( GUI) – Easy to use. For example for copying a file in window, only draggingthe file from source to destination will do.

2. Network capability – It is the capability of running the software with the users in the network fromthe server.

3. Web enabling - Availability of updates of the software to the users through internet connections. Forexample, users can have more images from Microsoft website via internet connections.

3.9 Advanced Programming Languages

C +

C+ is developed version of Language C. It is a structured language and supports features for developmentof structured programs. It is commonly used for system programming because of facility to have access tomachine level information through pointer, union etc.

Advantages of C+ :

• Program performance is very high in terms of speed of execution

• Structured Language

• Facilities of low level access

• Supports a good number of standard librabries etc.

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Disadvantages of C+ :

• The syntax sometimes becomes difficult to understand.

• The compiled program may have bugs and cause devastating results. A very unsafe programminglanguage.

• Understanding program logic from a program sometimes becomes difficult for debugging.

3.10 Visual Programming Language

Programming Languages which can use graphical features has become very popular these days fordeveloping user-friendly software.

3.10.1 Visual Basic (VB)

Visual Basic is a powerful language to develop GUI based programs in Microsoft Window Based system.It is evolved from the original BASIC language with GUI features making it a very simple and user friendlysoftware. It has also the special feature of integration with COM components written in other languages.COM components can also be linked to / embedded in application’s user interface.

Advantages :

• It is a simple Language

• It has comprehensive on-line help facilities.

• It supports GUI features

• It supports the facilities of integration of COM components written in other languages.

• It has Object Linking and Embedding ( OLE) facilities.

Disadvantages

• Programs written in VB is slower and less robust.

• VB programs has limited portability in Windows environment

• It has very limited access to library.

• OLE Object Linking and Embedding needs large amount of memory.

3.11 Object Oriented Programming

Object Oriented Programming is a way to organize data and code within a program. Object–orientedrepresentation is a powerful technique for dealing with a complex system. Three main features of Object–oriented programming are :

• Encapsulation

• Inheritance

• Polymorphism

Object : An object is a basis entity in a system. Every object consists of an attribute and one or moremethods. An attribute can be any property of the object. A unified entity or item can be viewed as an

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object. A person or city or cars can be viewed an object.

Class : Class refers to the definition of the object. Class helps in making the entire set of data and code intoa single user defined data type. Once a Class has been defined, a number of objects that belong to the sameclass can be created.

Attributes : Attributes characterizes the object and are identification criteria for an object. For exampleName, Phone number, Address are attributes for a person which is taken as an object.

Methods : Methods are functions and procedures that are used to perform actions related to the object.

3.12 Oracle

Oracle is a object Relational Database Management System ( ORDBMS). Object can be defined as re-usablesoftware codes. Oracle is based on a concept of ‘Client/Server Technology’. Under this technology, it hasto perform both the activities – database server and interaction with application client.

Tools of Oracle are :

• SQL * Plus – standard database language for storing and retrieving data

• PL/ SQL – Procedural Language/SQL.

• Forms – designing report

• Reports - output generation

Features of Oracle

• Ability of breaking table into smaller to reduce the risk of loss of data

• Provision of high data integrity

• Allows multi-user access to databases with explicit locking of data

• Ability to retrieve data spread over multiple tables etc.

3.13.1 Virus

It is a software which is directed to destroying software or information in a computer. It is generallycarried along with a program or data through a device without the knowledge of the user. The virussoftware is activated at encountering a particular situation either in terms of time or action. It gets multipliedautomatically that means it gets copied from one file to other in an attempt to make the damage wider.There are varieties virus and their actions are different. It destroys software, data file and even file allocationtable ( FAT).

The virus software was developed to interfere with free exchange of program in order to make the users tobuy software by paying due price for it to protect the interest of software developers. But its subsequentindiscriminate use has gone beyond ethics to affect the resources and free exchange of programs amongthe users.

3.13.2 Anti-virus

It is again a software to neutralize the effect of virus in the machine. It first detects the presence of a virusin a media and then removes the virus from it. Anti-virus program has a pre-defined ability to detect or

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correct against a range of viruses. Virus is like an infection and anti-virus is like a drug. There is tug of war

in the development process of virus and anti-virus and it has become a good source of business. There is aneed for continual up-gradation of anti-virus software to counteract virus. The best process of virus controlis to have sufficient preventive measures in importing software and files into a machine.

3.13.3 Virus Scanning

This is a process of continuous checking of possible encroachment of virus in a machine with the help of aset of anti-virus software called anti-virus tool-kit. Scanning provides safety to the software and data filesbut its negative effects are :

• It slows down the processing

• Requires continuous up-gradation of anti-virus software

• It involves cost

• It does not guarantee against the damage by new virus

Measures for Virus control : Following are the measures :

• Virus Scanning

• Up-gradation of anti-virus tool-kit

• Restriction to access by unreliable and unauthorized users

• Restriction of copying unauthorized software ( Game Software etc) which are major carriers of virus.

3.14 Application Software

An application software is developed to do data processing work according to the need of a particularapplication area, be it commercial or scientific or otherwise. An application program can be called ordedicated software for a particular work.

Major areas of Commercial Applications:-

• Payroll

• Inventory control

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• Financial accounting

• Sales accounting

• Production control

• Purchase accounting etc.

Example of Mathematical package for Scientific, Engineering applications :

• Statistical Analogy

• Multiple Regression Analysis

• Linear Programming

• Simulation and modelling etc.

Example of Special purpose Application Software

• Nursing Management in Hospital

• Air-line Reservation

• Railways Reservation etc.

Firmware : Firmware or micro-programs are a series of special program instructions. They basically dealwith low-level machine instructions to be carried out by different circuits. For this reasons, this type ofprograms are referred as firmware.

3.15 Applicability of Program Language

The applicability of the Program Language depends on the type of application. The features and codingstructure of a program language are more suitable to meet the requirement of some category of userswhich are given below :

For Scientific and Engineering application

BASIC

FORTRAN

ALGOL

APL

For Commercial Application

COBOL

RPG

BASIC

For General Purpose

PL/1

PASCAL

ADA

SystemDevelopment

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System Development Cycle

4.1 System Development

System development process involves three major steps – System study, system design and programdevelopment. In fact, a business system is not developed so simply. For a business system requirement hasto be studied, mode of processing has to be decided, system architecture has to planned, cost-benefit analysishas to be made, management approval is to be obtained and the following detailed steps are followed oneby one to make it full proof.

4.2 Development of Prototype

A prototype is smaller version of system in terms volume, complexity and cost. A prototype is developedas a pilot version to understand how the system will work and to assess what the things to be done to makethe system successful and to visualize what benefits will be accrued. Implementation of actual system is along term affairs. It helps to make decision making in giving signal for actual venturing into softwaredevelopment projects and what precautions are to be taken to make the project successful.

4.3 System Development Life Cycle

In fact system development involves seven stages as described below. These stages together is called SystemDevelopment Life Cycle ( SLDC). They are :

1. System Proposal

2. System Study

3. System Design

4. Program Development

5. System Implementation

6. System Evaluation & Review

7. System Maintenance

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4.4 System Development Team

For a large system, following members are involved in the system development. Their task and responsibilityare given below :

Step Person Responsible Responsibility

System Proposal Systems Analyst To study the requirement and prepare a proposal forclearance by the management.

Systems Study Systems Analyst Systems analysis of the existing system with Operationand MIS department employees and users andpreparation of System Analysis report for clearanceby User department and MIS department

System Design Systems Analyst Designing a system for checking by Project Manager/ Systems Manager and develop systems specifications

Program Programmers To develop programs according to systemsDevelopment specification and testing the programs

Systems Programmer & Implementation of the live dataImplementation Systems Analyst

System Evaluation MIS Manager & To assess the performance of the system and suggest& Review improvement

Systems Programmer To take care problems / new changes in the systemsMaintenance

Documentation Documentation To develop documents for usersExpert

4.5 System Proposal

In the System Proposal phase, the information requirement at different levels is assessed and the deficienciesand shortcomings in the existing system are studied to understand why it fails to deliver the desiredinformation / reports to satisfy the requirements. With careful consideration of the operational procedurefollowed and changes required, a new system is proposed. For the above proposal, the following detailedactivities are undertaken :

• Assessing enterprise’s information requirement

• Identifying the shortcoming in the existing system

• Prepare a plan for system management

• Work out a Feasibility Report

• Cost-Benefit Analysis

• Prepare a report on overall system objective and plan of implementation etc

4.5.1 Feasibility Study : This is part and parcel of system proposal. This step is only to in-depth evaluatethe details of processing involved and pros and cons of success factors for the proposed system. This stepinvolves the consideration for the following :

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• Size of scope of the project

• Systems Master Plan

• Risk involved etc

4.5.2 Analysis of existing system : The deficiencies in the existing system, need for possible improvementare studied and alternative sets of solution are devised. In this step, what is done is to see the existingsystem, their operational flow and the change in user’s requirement in the evolution process of InformationSystem.4.5.3 Study the flow of Information : The flow of information from users are to be analyzed in order tostrike a balance their action and their requirement. Analysis of I/O processing and its operational aspectsis done at this stage. The improvement in the data flow is suggested at this stage for better reportingsystem.

4.5.4 Scope of Solution : The aim is to freeze the requirement of the desired system after the studying thefollowing :

• Data to be processed

• Resources available

• Reports to be generated

• Time schedule to be maintained

• Reliability to be achieved

• Balancing the systems requirement and technical feasibility etc

4.5.5 Study of Technical feasibility

Technical feasibility means the consideration of the following technical factors of operation to match thesystems requirement :

• Hardware configuration

• Capability of the software

• Network requirement

• Skill of operation staff

• Data security features in the system

• Control system

4.5.6 Study of Operation Feasibility: Operation feasibility means the fulfillment of requirement of generationof data at designed speed, maintenance of flow of data, adherence of criteria of satisfying the users, clients,employees, vendors etc. The issues which are studied are :

• Resistance to change

• Management support

• Investment requirement etc

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4.5.7 Cost - Benefit Analysis: This is basically to justify the financial feasibility of a system. Before a newsystem is proposed, the benefits derived out of it should be weighted with the cost in it. Cost componentswhich should be covered in the analysis are :

• Investment requirement in hardware procurement /rent

• Cost of software development

• Maintenance of hardware

• Cost of creation/conversion of data files

• Manpower cost

• Cost of computer stationery including floppies, tape etc

• Training cost

• Cost of maintenance of software

• Overhead cost – power, air-conditioning, rent etc

Financial and non-financial benefits from the system are assessed.

Tangible benefits :

• Decrease in data processing cost

• Saving through better information system

Intangible benefits : It is difficult to estimate them. They are like:

• Increased operational efficiency

• Good working environment

• Better information system etc.

Cost and benefits are weighted and observations are made accordingly for proper decision making.

Report on overall System Objectives and Implementation Plan. Generally System objectives outline thefollowing :

• How the new system meets the changes in requirements

• Technological improvement envisaged

• Skill improvement plan for manpower

4.5.7 Implementation Plan : It indicates the benefits out of the system. Basic objective of this document is toplan the different activities and their time schedule :

• Investment Plan

• Software development schedule

• Personnel Resource Plan

• Managing the change

• Testing & implementation plan

• Training of the personnel etc.

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4.6 System Study

This stage focuses on understanding the requirement of the system in terms of reports and their frequencies.What are studied the flow of information, requirement of hardware and software and logic/rules involvedin processing of data at different stages. This step is also called System Analysis.

The details of analysis undertaken are as below :

• Analysis of the present system in terms of data flow, system logic and processing

• Understanding of shortcoming of the existing system in terms of speed of processing, flow ofinformation and forms of reports

• Understanding requirement change in data and data flow to meet the new requirements

• Understanding of System logic

• Understanding of Reporting requirement, volume and frequency

• Assessment of Integration possibility

• Devising mechanism of system performance improvement

Methodology adopted for system study :

• Examining the present manual, data forms and output report

• Discussion with knowledgeable persons from MIS and User department

• Analysis of input and output

• Preparation of questions on doubts for clarification

• Consideration/ examination of responses of questions

4.7 System Design

Focus in this stage is to design the data structure, file ( database), output. Specifications of the programsare developed in this stage to understand the inter-connections among different elements of different files(tables). In fact, System Study aims at understanding the whole physical system in an functional area(data, output etc) whereas System design aims at capturing the data in required specification so as to makethe processing possible to generate desired output/reports.

• Specification of data elements

• Specification of input form, records and files

• Specification of system input

• Development of system flow chart

• Development of program specification

• Determine file organisation & file layout

• Define reporting requirement, volume and frequency

• Development of operating specification

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• Development of control & audit specification

• Designing the Output reports

Tools used in systems design

• System Flow chart

• Case Tools

4.7.1 Systems Flow Chart : It is a tool to present data flow and process steps in graphical diagrams. Eachindividual symbol used in the diagram represents a definite element in describing the total flow in thesystem.

Symbol Used to indicate

offline Activity

Input document

Data flow process flow

Process step

Data storage

Decision box

4.7.2 Case Tool : CASE ( Computer Aided Software Engineering) : Case tools are used for having automaticdesigning assistance with the help of versatile system designing facilities available in them. They are usedin the following activities :

• Drawing flow diagram

• Designing file structure

• Defining Data Dictionary

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• Drawing forms

• Generating Report formats etc.

Example : System Flow Chart of Payroll ( Batch Processing):

Step 1

Step 2

Step 3

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Example : Inventory Control System ( On-line Processing):

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4.7.3 Designing the Input : Input documentation based on proper input designing and required input flowfor data for processing. Generally, input is prepared in a well designed form. The following points must betaken into consideration for designing input for data processing :

i) Contents – The data should fulfill the requirement of output reports.

ii) Format - The document should follow the natural sequence of data generation and the data size mustbe in conformity with the database design. Provision for separate box clearly labeled for separate item

iii) Timeliness – data must be prepared at right time to facilitate the schedule of data processing.

iv) Authentication – Input data must be authenticated by the authorised person.

v) Media – Creation of data file should be made on a media like floppy disk, tape or hard disk accordingto the design specification of the system.

vi) Volume – In case of huge volume of data, for data entry key-to-disk or key-to-tape system may befollowed and the same may be processed in batch.

4.7.4 Codification of data : Codification is a system of convenient representation of data. Code is scientificrepresentation of actual data in short form. Codification is adopted to represent long data by shorter name,to reduce the error in data entry, to classify the transactions easily and making data processing efficient.

For developing a scientific codification to take full advantage of it, the following points should be takencare :

i) Uniqueness : Unique code for each separate type of data item

ii) Convenience : Codes are generally comprised of numerical digits but alphabets may be used in thebeginning to have more logical classification. Logic of codification should be simple and easyunderstandable by the users.

iii) Sub-grouping : Within the code, a group of characters are taken together as sub-code. This is to simplifycodification and classification.

iv) Correction mechanism : Sometimes, check digit is also used to ensure accuracy in data entry.

v) Flexibility : Coding system should be flexible enough to ensure smooth incorporation of futuremodifications and accommodation of possible additional sub-groups due to changes in the system. Inother words, future modification should not invalidate code structure.

Example : Employee Code helps to identify an employee easily. There is no need of entering his/her longname. Short Employee code serve the purpose. Similarly, the common codification are used for AccountsCode, Party Code, Item Code etc.

Accounts Code – for Accounts Heads in Accounting System

Party Code – for Party Name for Sundry Debtors, Sundry Creditors etc

Item Code – for item in Inventory Management System etc

Advantages of Codification :

• Shorter code in place of actual data reduces errors in data entry.

• Codes are used to classify the transactions easily

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• Sorting or merging by code is a scientific system step in data processing. Without Code data processingwill be cumbersome.

• Codification helps report generation to be efficient

4.7.5 Designing Source Documents: While designing source documents, the following points should beconsidered :

• Documents should be simple to understand

• Each document should have clear title.

• Each type of document should have a serial number.

• Different types of documents should be given different colour codes.

• Items of data should be put in the order of logically sequence.

• Layout of documents should be in standard size.

• One source document should be used once only to feed data.

4.7.6 Consideration in changing in existing source documents : The following precautions must be taken inchanging the existing source documents for smooth change over :

• Changes should be kept as minimum as possible so that data preparation can continue without muchhazards.

• Changes in new source documents should be limited to the requirements of the new system.

• Deficiencies of the existing source documents must be analyzed first and new documents should aimat fulfilling those.

• Data contents of the new source documents should be available at one place.

• Staff may be trained to equip them to efficiently handle the new source documents.

4.7.7 Designing the Output Reports : The following important factors must be taken into considerationfor designing the output report :

i) Contents – set of information to satisfy the users of the report

ii) Format – to place the information in logical sequence to increase the readability and effectiveness ofthe report. In case of graphical presentation of the output, the selection of suitable format is important.

iii) Media – In earlier days all output reports used to be printed on paper for the users. Now with theadvantage of on-line control, for example in banking services, or in case of real-time system likeRailways reservation system many output requirements are on VDU. Thus, media of output reportshas become an important consideration.

iv) Frequency – frequency of report depends on the need of the users.

v) Volume – Volume of report is a consideration. A voluminous report can not be on VDU.

vi) Usefulness - The purpose of the report may be for documentation ( preservation of records ), controlinformation, interactive decision making etc. This aspect is considered for designing.

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4.8 Program Development

Program development means writing programs according to program specification. This is also calledcoding.

The consideration for program writing are :

• Tools for developing program logic

• Programming Language

• Input design

• Files (tables) to be handled and their design

• Output / reports design

Program Development has the following steps to be followed sequentially :

i) Development of program logic with the tools

ii) Selection of Programming Language

iii) Writing the Program ( Coding)

iv) Debugging the Program

v) Testing the Program with test data

vi) Implementation with live data

vii) Program Documentation

viii) Maintenance of the program

4.8.1 Tools for development of Program Logic : The tools available development of logic of the programare:

i) Decision Table ii ) Flow Charts and iii) Algorithm.

Decision Table

It is a tabular representation of program logic showing the actions against the conditions. The componentsof a decision table are :

i) Condition Stabs - different condition entries in different rows ( upper)

ii) Action Stabs - different action entries in different rows ( lower )

iii ) Rules - Rules are unique combinations of conditions i.e. combination of ‘Y’ and ‘N’where ‘Y’ means ‘yes’ and ‘N’ means ‘no’ No of rules = 2 x Number ofconditions

iv) Condition entries - under the rules, the combination of conditions are entered.

v) Action entries - Against the each condition entries, ‘X’ has to be entered corresponding to theaction to indicate ‘yes’ according to the logic of the problem.

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Example 1 : ABC Co extends special discount to its distributors who have business relation for more than7 years or have business relation more than 3 years and volume of business is more than 10 lakhs. Draw theDecision Table.

Answer :

Rules

1 2 3 4 5 6 7 8

CONDITION C1 : more than 7 years Y Y Y Y N N N Nbusiness relation

C2: More than 3 years Y Y N N Y Y N NBusiness relation

C3: business volume Y N Y N Y N Y NMore than 10 lakhs

ACTION A1: Discount facilityA2: No Discount facility X X X X X

X X X

Example 2: A manufacturer sells his products to three different types of customers with different discountrates based on order vales as :

Customer Commission % on Order Value

Less than Rs 7500 Rs 7500 to less than Rs 15000 Rs 15000 and more

Retailer 5 8 10

Distributor 8 10 12.5

Govt. party 8 8 8

Draw the decision table.

Answer:

Rules

1 2 3 4 5 6 7 8

CONDI-TION C1 : Order < Rs 7500 Y Y N N N NC2 : Order > = Rs 7500 N N Y Y N N

And < Rs 15000C3 : Order - More than N N N N Y Y

Rs 15000

C4 : Customer : Retailer Y N Y N Y N NC5 : Customer : Distributor N Y N Y N Y NC6 : Customer : Govt party N N N N N N Y

ACTION A1 : Discount 5% XA2 : Discount 8% X X XA3 : Discount 10% X XA4 : Discount 12.5 % X

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4.8.2 Flow Chart: Flow Chart is the diagrammatic representation of the algorithm i.e. program logic. Ituses a unique set of symbols to describe the conditions and actions.

Technique of in drawing flow chart

• Starting with the input for main decision factor

• Putting the condition in the decision box

• Branching the condition into different path of decision box

• Branching should be done without ambiguity

• To remove ambiguity, more than one decision box may be used

• Writing the statements under each branch

• Avoiding crosses in the flow chart

• Using connectors to reduce the number of flow lines, if required.

Advantages of Flow Charts :

• Logical representation of problem steps

• Flow chart helps to make the complex logic simpler

• A visual aid in conceptualization of the problem

• It is tool for efficient programming

• Helps in debugging

• Support program documentation

Symbols used in flow chart

Symbol Used To Indicate

START / STOP

INPUT / OUTPUT

PROCESSING

FLOW LINE

DECISION BOX

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Start

Read Purchase Order Value

PO Value < 100000

Discount = PO Value x 0.05

PO Value < 200000



Stop

Yes

Yes

No

Example 1 :

In a company A, discount on purchase is given on the basis of different slabs in the value of the order asgiven below :

Purchase value Discount %

Below Rs 1,00,0000 5%

Rs 1,00,000 – 1,99,000 7.5 %

Rs 2,00,000 & above 10%

Flow chart for calculation of discount :

4.8.3 Algorithm : Algorithm is the logical steps of problem decision. A problem is divided into distinctlogical conditions and actions under each conditions are framed. This is a tool for developing a programfor a given problem.

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Steps involved in developing algorithm :

• Understanding the problem

• Picking up the parameters involved in the problem

• Making combinations of parametric values to arrive at distinct logical sub-conditions without ambiguity

• Framing action path corresponding to each sub-condition

• Testing the algorithm with test data and checking the output

Advantages of Algorithm

• It helps the programmer to understand the problem thoroughly by breaking the logical conditions

• It brings solution step by step

• It is easy to debug the program with the help of algorithm

• It is independent of programming language

4.8.4 Program Life Cycle : Program Development follows the following seven stages :

4.8.4.1 Program Analysis – This is to understand the users requirement and logic of the business processto be taken into consideration. The programmer determines the requirement of the system in terms ofinput data handling, output reports generation and logic involved.

4.8.4.2 Program Design - In this step, the program is divided into different modules. For each module,algorithm and flow charts are developed to understand how program will work. A document is preparedwhich is called program specification. Programmer has to use the tools like file layout, input design, outputreport format and outlines of program logic. All these taken together form the program specification.

4.8.4.3 Program Coding : In this step, program is written. First, language of the program is selected andthen coding of the program is done. The Programming language depends on the following criteria :

• Application area

• Data Structure

• Program logic complexity

• Skill of the programmer etc

4.8.4.4 Program Debugging : Debugging a program means eliminating errors in the program. Errors maybe of two types - a) Syntax error and b) logical errors

Syntax errors : These are errors due to mistake in following the syntactical structure of the language inwhich the program is developed or there may be spelling mistake in the use of words and instructioncodes. These errors are detected at the time of compilation with the help of a compiler. Unless the syntaxerrors are corrected, object program will not be generated.

Logical Errors : These are errors in program logic. These are generally detected at the time of execution ortesting the program. Due to logical errors, program may fail to run or will generate wrong output report.

Compilation of Program – Once the free from error, the program has to be compiled with the help ofcompiler of the language chosen. This compiled program is run.

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4.4.8.5 Program Testing : Program testing is part of system implementation. It is done with the help of testdata to test the correctness of program logic in terms of data manipulation, file handling and outputgeneration. This testing is done to make the actual implementation with live data simpler.

Test Data Preparation – Test data is to be prepared to check the correctness of the program.

The following excises are undertaken in this stage :

• Creation of realistic test data very similar to live data with proper boundary values

• Processing the test data with the help of new software

• Examining the correctness of the reports generated

• Verifying the status of updated files ( table) to check correctness

• Identifying the errors and their causes

• Correction in the programs, wherever necessary etc

Debugging Testing

Process of eliminating errors Process of validating the logic of the programIt is after the program is written. It is after debuggingIt is done with help of compiler or interpreter. It is done with test data.It is done to eliminate syntax errors. It is done to remove logical errors.

After the software is successfully tested with test data, the next step of effective way of testing the systemis to make a parallel run of old and new software with the help of same set of live data. The results of boththe system is compared to identify the variations. The parallel run must be completed with live data for aconsiderable period and all reports must be checked to ensure complete correctness of the system.

Difference Between Debugging and testing :

4.8.4.6 Program Documentation – Writing narrative details of the program. Documentation for program isknown as Programming Manual. It gives logic of each program with the following details :

• Algorithm and flow-chart

• Program specification

• Source code

• Output report

4.8.4.7 Program Maintenance – This is required to accommodate the changes in the system. Properdocumentation of all these changes must be maintained.

4.8.5 Quality of Program : Quality of program depends the following criteria :

• Accuracy - with no logical flaw.

• Reliability – giving consistent result.

• Readability – flow of logic is understandable

• Efficiency - Performance should remain good even with high volume of input data

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• Maintainability – program with simple logical flow can easily be corrected by somebody other thanits own developer.

4.8.6 Structured Programming

Structured programming means the art of developing programs in a structured fashion to make it readableand maintainable. The rules of structured programming are as follows :

A. Logical flow will have one entry and one exit.

B. Three basic structure :

� Sequence of execution ( DO)

� If —— GOTO

� If —— Then —— Else

C. No haphazard use of branching using GOTO

D. Top-down or bottom-up approach

4.9 System Implementation

At this stage, both hardware and software are put in operation. In the system implementation phase, themost critical task is the preparing work force. People resist change. Forceful voices ( Union, employees,Supplier, Customer etc) are to be identified. They are to be made understand that there is no threat. Prospect,advantages and improvements are to be explained. Organizational readiness to accept the change is to beassessed.

4.9.1 Activities in System Implementation

• Adoption of new technology

• Procurement and installation of hardware

• Development of communication system

• Establishment of control measures

• Fault tolerant system development

• Development of procedure for system administration

• Loading the System in the machine

• Conversion of files

• Establish the new system of data flow and capture data

• Process data and verify the reports.

• Establish control procedures – both preventive and corrective

• Make provision for sufficient flexibility

• Training to operation staff

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• Establishment of system of communication between operators and users

• Scheduling for report generation etc.

Conversion of files – transfer of all files, databases from the old system to new system. Normally file/database design in the new system will differ from that of old one. To make things operational, the conversionis required.

4.9.2 Requirements for System Change Over: Change over means the switching over from one system toother. This becomes necessary along with change in environment and improvement in information systemaccording to the requirement of development in management system. The change over involves a verygreat complexity in terms of up-gradation of hardware, software, skill requirement and operation. Followingobviously requirement in terms changes in the system management are :

i) Conversion of files : New system will have change in the file design. Existing file must be converted inthe new file structure in order to preserve the existing data and maintaining continuity of the processing.In case of newly design system, adoption database management system is in high proportion. The olddata files in conventional file design need to converted in database.

ii) Skill improvement : Skill requirement for operation of the newly designed system is expected to bedifferent. The users, system administrator, operator need to be trained to understand the change andtechnical matters involved in handling the new system. The thorough training on the new hardwareand software features are essential for smooth running of the system. For example, if the new systemis on DBMS the operating staff needs training on it, users need training to how to use it as on-linesystem and to have the information on interactive mode.

iii) Up-gradation of hardware : Old Computer Systems are to be replaced according to the requirementof the new system. If the new system is on net-working environment, the required hardware devicesare to be arranged.

iv) Operation Scheduling : According to the requirement, the job allocation and responsibility has to bechanged. The required security features in the data files and the system has to be established.

v) Conversion of the Software : Conversion of the software involves the testing the new software withthe help of converted data-files ( databases). The data-processing staff should be involved in processingthorough understanding of sequence of operation, security features, precautions to taken to preventdamages etc.

4.9.3 Different approaches for change over

1. Parallel Run – Both the old system and the new system are carried on till a point is reached when thenew system is streamlined to take over the processing in its full form. This approach has the provisionto have comparison in the reports, their correctness and desired efficiency in the new system.

2. Immediate change over - This approach favours the implementation of the new system straight wayunderstanding the obvious problems in implementation. The approach gives more stress on solvingthe problems in the new system and taking care of the further requirements in continuous developmentprocess. This approach is adopted when there is a thorough change in the system and there is littlescope in making a one to one comparison of the outputs of the old system and the new system.

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3. Phase Wise Conversion - This approach is adopted only to avoid sudden load of full conversionprocess when the system is extensively widespread and are of broken in modules. Sometimes, thisapproach is adopted to avoid huge investment requirement in total change of hardware in one go.Another advantage in it is training of people in the new environment batch by batch with dislodgingthe whole old system at one point of time. This is a process of progressive changeover module bymodule and then testing of interconnectivity will be done in the last phase. This process is sometimesproves to safe and effective.

4.10 System Evaluation & Review

In the system evaluation phase, critical evaluation is made on whether the system performs in the expectedlevel and targeted cost-benefit is achieved. The scope of improvement is also identified. The followingexercises are undertaken in this stage :

• Review the users’ initial feedback

• Evaluation of performance of system in terms of speed, reliability and timeliness of reports

• Storage and backup provisions

• Cost Benefit Analysis

• Evaluation of internal controls

• Evolving measures for continuous improvement etc

4.11 System Maintenance

System Maintenance is undertaken to take care of unforeseen errors in the programs or due to bug in thedesign. It also undertake continuous improvement in the system as envisaged in the Evaluation and Reviewstage. The activities in this stage are :

• To constantly watch the performance of the system

• To identify the points of modifications in the system to achieve higher efficiency

• To suggest better control mechanism to avoid risk

• To see the behaviour of the some programs in handling critical data sets

• To develop procedure for recovery from system failure

• To follow standard procedure for back-up of databases

• Standardisation of procedure of maintenance of hardware etc

Maintenance of Maintenance records is very essential to keep track of the maintenance work undertakenand to understand the performance of the system :

• Problems faced in the system and their frequency

• Quality of maintenance work

• Date of modification and nature of modification

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4.12 System Documentation

System documentation is an important issue for the user. It provides detailed information of system tomake users conversant on system with details of steps involved in the systems. It is an aid to users inrespect of the following :

• Material for training

• A mean for implementation, maintenance and modification of system

• Guidance and operating instructions to users and operators

Contents of System Documentation

• Program description

• System description

• Data and file specification

• Input/output specification

• System error controls

• System test plan

• Program specification

• Program flow chart

• List of programs

• Control and audit trail

• Glossary & terms used.

Contents of Operation Manuals

• Program description

• Input/output specification

• Operating instruction of programs

• Provision of modification, up-dations and deletion of records in files/databases

• Program controls

• Output specification

• Precautions to be followed

Contents of User’s manual

• Purpose of the system

• How to prepare source documents

• Design of input forms

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• List of codes for various data fields

• Output specification

• Handling errors in source documents

• User’s control over schedule of processing

4.13 System Development Controls

The following controls in essential during system development :

• Proper authorisation of document preparation for input data

• Documentation of Systems specification in System Manual

• Adherence of standard system of testing procedure

• Creation of password for authorised access at different levels of operation

• Development of system of maintenance of log book for system usage for recovery management incase of damage etc

4.14 Software Development Cost

• Manpower cost or consultancy fees for system study, design, development and implementation

• Cost of creation/conversion of files

• Opportunity cost of hardware/rent

• Cost of language software or data base

• Training cost

• Cost of electricity, stationery

• Other overhead costs etc

4.15 Operating Cost

• Cost of Manpower

• Cost of computer stationery – ribbons, floppy, magnetic tape etc

• Maintenance cost

• Insurance cost

• Cost of electricity

• Depreciation

• Other overhead cost etc

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4.16 Readymade Software

Ready made software means an application package developed with the standard features. As it forapplication by wide range of customers, it is developed with adequate flexibility. Cost is comparativelylow. Upgraded versions are available with more features at a low additional price.

Before selection of a ready made software, the following points must be taken into consideration :

• To what extent it satisfies the requirements

• To what extent the package can be modified to suit the requirement

• Input/output formats – whether they are acceptable

• Package constraints

• Control system

• Quality of documentation

• Training provisions

• Costs

• Maintenance provisions and costs

• Warranty period etc.

Implementation process in ready made software :

• Hardware procurement according to requirement of the software

• Codification and training of personnel and implementation are of standardised.

• There may be a need to change the physical application system according to the codification and theprocessing need of the system.

Advantages of Ready Made Software :

• Rapid Implementation

• Standard and established package covering a wide range of options

• Availability of standard manual

• Low risk of failure

• Training provision from vendor

• User-friendly

• Cost is low

Disadvantage of ready made software :

• Does not exactly satisfy the requirement of organisation

• No flexibility to accommodate any special features of system according to the need of the organisation

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• The physical system and codification may be required to change to generate input documents accordingto the requirement of software .

• Hardware configuration according to requirement of software

4.17 Tailor made Software

Tailor made software means a software developed according to the need of a particular organisation. Thiscan be done after following the total system development cycle for a system. The advantages are takingcare of full requirement of the system with provision of change in future to accommodate new developments.The disadvantages are the long implementation time due to full software development cycle followed andthe higher cost involvement.

Implementation process :

1. Following Software Development Cycle

2. Procurement of hardware devices needed

Training of the operating personnel and users.

Advantages and disadvantages of ready made software and tailor made software:

Ready made software Tailor made software

Standard system Systems according to the need of theorganization

Generally user-friendly - easy to implement System development process has to passthrough SDLC. Naturally implementation timeis longer.

Hardware must be according to the minimum Software may be developed exactly to matchconfiguration requirement of the software. the hardware configuration planned.

Systems requirement is pre-defined and rigid. The system will have enough flexibility toIt may require to bring change in physical accommodate future changessystem and data flow

Training for operation of a standard system Training for a non-standard is difficult. As theis easy. Training is significant as the list of software is according to the existing physicaldo’s and don’t’s are big. system, adoptability will be higher.

Standard and comprehensive Manuals are Users manual has to be developed by thesupplied by the vendor. software development team.

Cost of software is low. Development cost is high.

DataManagement

andData

Processing

STUDY NOTE - 5

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Data Management and Data Processing

5.1 File

In data processing environment, file organization plays a very important role. Depending on the type ofapplication, procedure followed in the processing and the mode of processing followed for informationgeneration, file structure is designed.

A file is a collection of records. A record is made up of related fields in a particular order. The various fieldsconsist of group of characters. A character may be alphabetic, or numeric, alpha numeric or a specialcharacter.

A data field represents a specific data element. A record is designed according to the need of the informationsystem and fields are arranged in an data file in some pre-defined order.

Example- Employee Data file in Payroll System contains records of employees. A record consists on differentfield giving different details of the employee concern. For example, a record of employee file contains thefollowing fields:

Field Name Field Description Size Type

1. EMPNO Employee Number 5 A (Alpha Numeric)2. ENAME Employee Name 30 A (Alpha Numeric)3. DT-BRTH Date of birth 6 N (Numeric)4. DT-JOIN Date of joining 6 N (Numeric)5. ESEX Sex 1 A (Alpha Numeric)6. DESG Designation 10 A ( Alpha Numeric)7. BP Basic Pay 4 + 2 N ( Numeric)

Etc

Example - Actual data i.e. employee number for an employee is E0750. It is a alpha-numeric field.

5.2 Data Structure

Fixed length file – Record length is fixed.

Valuable length file – Record length is variable.

Key Field- A record is identified by its key field. This key field is used for searching a particular record forprocessing, display, editing etc.

5.3 Types of Files

5.3.1 Master File - Master file contains the information which are of a permanent in nature. In fact, in dataprocessing environment, master files contain fixed information. These information are changed only byperiodic updation, as and when the same is changed.

For example, an employee’s master file contains the fixed information of employees. Normally basic pay,designation, department is fixed information. In case of promotion or transfer these information may change.In case of joining or retirement or resignation, file will be changed with additional record or deletion ofrecord. Similarly, an accounts master file where the balances of accounts heads are maintained, gets changeddue to periodic updation.

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5.3.2 Transaction File:- Transaction files are the files in which transactions in a particular application arerecorded. In payroll system, monthly attendance file is a transaction file where the records of attendancesof employees during a particular month are entered. Similarly, for an inventory control system, all issuesof stock items are entered from the issue notes during a particular period to create transaction file onissues.

5.3.3 Work File:- These are basically intermediate files created during processing for an application area.These work files are preserved for the sake of security or to keep provision for reprocessing in case ofdamage or loss of output file at a particular stage. When a system is designed, the designer specifies whatwork files are to be preserved for how long and the procedure to be followed to use work file in case ofneed of recovery.

5.4 File Organization

Organisation of a file is done with respect to the mode of access to a file is required during processing.Following are the types of File Organisation :

1. Sequential File

2. Random ( Direct) File

3. Indexed Sequential File ( ISAM)

5.4.1 Sequential File - Under this type of file organisation, records are placed one after another and the filecan be accessed only sequentially one by another. The file may be sorted on the desired field to make dataprocessing more effective and meaningful.

Advantage of sequential file :

i) Simple

ii) All machines and software have the provision to support sequential file.

iii) Easy and economic to create file

iv) Maintenance of back up is simple.

v) Easy to reconstruct

Disadvantage of sequential file :

i) Sorting required before processing.

ii) Whole file is to be processed even when activity rate is low.

iii) Same data field is stored in different data files creating data redundancy.

iv) File handling involves intricacy in logic.

5.4.2 Randon (Direct) File – Under this type of file organisation, data are stored sequentially on the valueof the key field irrespective of order of creation of records. Provision of access to a record of having aparticular key value is efficient. The access is not done sequentially rater direct. This is why, file organisationis called direct or random. The position of the record with a particular value of key field is obtained byarithmetic calculation. This helps to have direct access to the record.

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Advantage of Random file :

i) Easy and fast access to the desired record makes processing efficient.

ii) Sorting of file is not required.

iii) File handing in data processing is simple.

iv) Updation of records is simpler.

v) It allows sequential access for processing.

vi) Most suitable for interactive and online applications.

Disadvantage of Random file :

i) Expensive hardware and software support are required.

ii) Storage space requirement is high.

iii) Addition and deletion of records in the file is difficult.

iv) Backup must be made with special security measures.

v) Programming logic is complex.

5.4.3 Indexed Sequential File – Under this type of organisation, records are kept sequentially and index isused to have faster access to the desired record. This means this type of organisation has the advantages ofboth sequential organisation and random organization. The index of key field maintain the logical recordkey and physical record key. For searching a record, this type of organisation follows a similar approach ofsearching a word from the dictionary.

Advantage of ISAM file :

i) File creation is simple.

ii) File access is moderately fast.

iii) File handling in data process is simple.

Disadvantages of ISAM file

i) Expensive hardware and software provision are required.

ii) Storage space requirement is high

Slow retrieval of records.

5.4.4 Comparison for dofferent file organisation

Sequential Random ISAM

Access Serial Random / sequential Random / sequentiaPossible media Magnetic tape / disk Magnetic disk Magnetic diskNormal uses Transaction file Master file Master file

Work fileMaster file

File design Simple Complex ComplexRecord length Variable / fixed Fixed FixedSoftware cost No special software Special and high Special and high

cost software cost software costUpdation of file Slow Fast FastBack up Easy With proper security With proper security

measures measures

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5.4.5 Criteria for File Organization:File organization in a system depends on the following criteria:

i) File Activities – It indicates the number of successful attempts of having access to records in the filesduring the course of a process of operation. File activity in mathematical terms is a ratio as givenunder :

File activity = No. of records accessed / No of records in the file

If file activity is high, direct or ISAM file organization is better.

ii) File volatility – Volatility is the rate of change in the records in the file. Generally, volatility is consideredfor master file.

iii) File Size – large file handling is easy in sequential mode. In earlier days, when hardware cost washigh, the preference was for sequential file. Now, the need of processing is considered to be mostimportant.

iv) Back up – Back up provision for sequential file is easy . Back up of sequential file can be kept both inmagnetic tape or hard disk.

v) Batch processing – Generally in case of batch processing, both master file and transaction files arecreated following sequential file organisation and processing is done at a particular point of time. Thebatch processing, file activity is high but file volatility is low.

vi) On-line processing – Generally on line processing involves high file activity and requires fast accessprovision to make processing faster. In that case, Random or ISAM organisation is suitable.

5.5 Data base

A data base of a collection of inter-related data with controlled redundancy to serve one or more applicationsin fulfilling their need. Database is used by an organisation to store its data from different operationalareas so that they can be shared by each operation collectively. Use of database has become very popularover the conventional file system for two reasons - integrated information and its security features.

Characteristics of data base.

• Information Sharing – Data is shared by different applications

• Controlled redundancy - In a non-database system, each application has its own files. In a database,duplicate data is eliminated and thus redundancy is controlled.

• Consistency – Data set from database (i.e. from one place) ensures better reliability of data instead ofcollection of different files.

• Security – Unauthorised access to data base is restricted by use of password.

• Integrity – Data is kept in integrated fashion and updation at one place ensures high integrity

• Persistence – Data in the database exist permanently.

• Independence – The three levels in the schema ( internal, conceptual and external ) should beindependent of each other so that the changes in one level does not affect the other.

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5.6 Database Management System (DBMS)

DBMS means a specialized software to create and modify a database. A DBMS provides a mechanism ofsharing and manipulating by properly authorized multiple users without compromising the security andintegrity of data. DBMS can make several approach to manage the data.

DBMS Offers the following services :

i) Data Definition – defining the data structure for storage of data

ii) Data Maintenance – Checking each records whether all information about one particular item. Forexample, to see whether in an employee table, all information about employee are there- like name,department, designation, salary etc

iii) Data Manipulation i.e.

• Insertion

• Updation

• Deletion

• Sorting

DBMS Software – A software supports the following :

• Data Definition Language’

• Data Manipulation Language

• Query Language

• Report Generator

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Data Definition Language ( DDL) – A language having sets of statements to define, maintain and a schemaobject.

Data Manipulation Language ( DML) - A language having sets of statements to manipulate data in database.For example, insertion, deletion, updation of rows in a table and query on data from database.

DML compiler – It compiles DML statements into an application program.

Query Language - A language which enables an user to ask specific query to the database.

Report Generator – It enables the users to design and generate reports including graphical representation.

File Management – System of maintenance of data according to the design of the database.

Authorization to database Access – Database Administrator authorises different types of users differenttypes of access to the database to control security of database.

A Database Management System has different modules which takes care of different functions of a database.They are :

Module Function

File Manager Allocation of space on disk storage and store data structure

Database Manager Interface between application programs and database

Query Processor To Translate a query language to low level instructions to help processingof users requests

DML compiler Conversion of DML statements into application program

DDL Compiler Conversion of DDL statements into a table

Advantages of DBMS :

• Reduces data redundancy

• Ensures data consistency

• Provides check for data integrity

• Maintain in-built data security

• Offers flexibility to change data structure in case changed requirement

• Facilitates storage of structured information

• Provides concurrent access to multi-users.

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5.7 DBMS System Structure

Relationship among data :

1. One-to-One

2. One to Many

3. Many to Many

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Different models of DBMS :

1. Hierarchical database

2. Network database

3. Relational database

5.7.1 Hierarchical Database:

The basic concept in the hierarchical database is that the data is organized on the basis of a chart of definiterelation between different levels of data.

Main features of hierarchical database structure are:

i) Parent record will have some child records (all linked with it).

ii) A record becomes a child when it corresponds to its higher level record.

iii) The same record may be a parent record in relation to its lower level (child record).

iv) A ‘child’ has one parent but a parent may have many child.

v) Between two consecutive levels of records, the relation is unique.

vi) It can not represent ‘many to many’ relationship.

5.7.2 Network Database:

In this type of database, each mode may have convection with many other modes showing the structure tobe similar to be network. Nodes are connected in a multi-dimensional manner. When the relation amongthe modes are not as simple as in the hierarchical model. This model has many flexibilities and is suitablefor a complex relation among the modes. The main advantage of this model is that the database canaccommodate a number of inter-related records with relational path to be a network standard. Thecomplexity of the structure is its problem to have wide acceptance.

The relationship is many to many. This is shown with the help of a diagram where three products are soldto different customers.

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5.7.3 Relational Database ( RDBMS):

Most widely used database model. Under this models, tables are formed with several closely related dataelements and relation is established among the tables. No two rows in table can be identical. Relationshipbetween two tables are established at column levels and the related data elements from two tables fromdifferent rows can be linked. The concept is very popular in the sense it is almost similar to conventionalfile system. The relational database software has a good provision for linking relationship among differenttables and establishing security features for maintaining integrity of data.

Vendor Table

11 Company name Company Address

V1 Vendor 1 Address of Vendor 1




Item Table

Item Code Item Description

11 Item 1

12 Item 2

13 Item 3

14 Item 4

Purchase Vendor Item code Item Qty Purchase OrderOrder No. No. Value (Rs.)

P1 V1 I1 200 3000

P1 V2 I1 500 7500

P2 V2 I2 600 6000

P3 V4 I3 900 18000

Among the three above tables, relationship are established by Vendor No. and item code.

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5.8 Normalisation

Normalisation is a process of organising data in a database. The fundamental principle of normalisation isthat the same data should not be stores in more than one place. The basic aim of normalisation in datastructure is to eliminate redundancy and inconsistent dependency. The design of database with greaterlevel on normalisation provides better efficiency.

5.9 Code’s Rules

Dr E F Codd first introduced relational database model in 1970. Codd’s Rules are designed for stablestructure for Relational Database. Rules are :

1. Information Rule – All information is explicitly and logically represented in tables as data values.

2. Guaranteed Access Rule – Every item of data must be logically addressable with the help of tablename, primary key value and column name.

3. Systematic Treatment of Null Values Rule – RDBMS must support null values to represent missingor inappropriate information.

4. Database Description Rule – Description of database should follow the same logical structure withwhich the data are defined in RDBMS.

5. Comprehensive Data Sub Language Rule – The system must support the following :

• Data Degfinition

• View Definition

• Data Manipulation

• Integrity Constraints

• Authorisation

• Transaction management operation

6. View Updating Rule – All views that are theoretically updateable must be updateable by the system.

7. Insert and Update Rule – For all Insertion and updation, a single operand must hold good.

8. Physical Independence Rule – When any changes are made in storage, the application programs mustremain unimpaired.

9. Logical Data Independence Rule – Changes in data should not affect user’s ability to work with thedata.

10. Integrity Independence Rule - Integrity constraints should be stored in the system catalog or in thedatabase as a table.

11. Distribution Rule – The system must be able to access or manipulate the data that is distributed inother systems.

12. Non-subversion Rule – Integrity constraints defined by user must not be by-passed by RDBMS.

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5.10 DATA PROCESSING

Data Processing – Data processing means conversion of raw data to meaningful information.

DATAPROCESSING DATA INFORMATION

Raw data means the data generated from application system. It is called raw as a single piece of data orsome records on transaction do not provide sufficient means in decision making process. Data processinginvolves calculation or manipulation with those data in terms of sorting/merging, comparing and deductinga result whereby it makes some sense in the decision making process. In other way Data Processing can bedefined as a process which generates output from some input in the

DATAPROCESSING INPUT OUTPUT

5.11 Data Processing Activities

5.11.1 Data Entry: Data entry generally means entering the transaction data from source documents eitheroff line or on-line. Off-line means an environment where the validation of transaction data is not done withthe help of corresponding master file and file is supposed to be created in the chronological order of thesource documents. On-line means an environment where validity of existence of key records or logicalcorrectness checking of other fields are being done.

5.11.2 Transaction file creation: This is done from the file created by data entry process with the followingsteps:

• Validation of data

• Checking and Editing

• Sorting/Merging on the keys required for processing

5.11.3 Sort & Merge : Sort and merge are two operations which are very common in data processing.Sorting means reorganizing the file in terms of sequence of a field or fields ( sort keys) whereas mergingmeans combining two or more files ( having same specifications) on some fields. This is done with the helpof a software which takes in an input file, specification of desired fields on sorting is required and name ofthe output file. If the input files are more than one and they are sorted on some fields to produce oneoutput file, then it is merging the input files.

5.11.4 Updation: Updation means modifying the old information with current information – addition ofrecords (insertion) – Deletion of records – modification of records. This is repaired when the informationwas incorrect and detected at the time of editing or otherwise. Another situation is very common when theinformation needs modification with the effect of normal business transaction. This updation may be on-line or batch mode. In a batch mode, updation takes place with the help of a transaction file. With the helpof a program, the records are changed. Now a days with the change of technology, updation on on-line isbeing done to have instant result.

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In a Railway Reservation system, updation is on-line. In case of reservation or cancellation of ticket, updationtakes place immediately and output of ticket comes out instantly.

5.12 Factors For Decision On Mode Of Data Processing

The following are the factors to decide on the mode of data processing:

i) Volume of transactions and frequency of transactions

ii) Organizational structure – Organizational structure decides the points of sources of data, processingneed, method of decision making etc

iii) Information requirement and its schedule – Process of decision making, type of data and impact ofdecisions on business process, control mechanism decides on information requirement, report structureand their frequencies etc.

iv) Computer system configuration – System of Computer configuration includes hardware configurationin terms of speed of processing and storage space, system software available in terms of operatingsystem and language etc to determine the capability in order to select appropriate mode of processing.

v) Communication network requirement – Organisational structure and requirement of informationcommunication or integration decides the type of network topology required to suit the decisionmaking processing and information storage and management.

vi) Capability of software – The software capability is to judged to match the above requirements and toselect appropriate data processing mode.

5.13 Advantage of Data Processing

The following are the main advantages of data processing :

1. Speed: Speed of the data processing is the main advantage.

2. Volume of Job : A huge volume of data can be handled in data processing environment with ease.

3. Accuracy : Once software for an application area is tested and made error-free, accurate results can beensured, irrespective of volume of transaction.

4. Reliability: The computer is much more reliable than human being in the sense it does not have thefatigue to do the repetitive jobs.

5. Consistency: In case of manual processing, the steps of calculation or approach may differ with thechange of person responsible for the job and results may differ. In case of computer processing resultwill have steady consistency.

6. Timeliness: Schedule for processing can be planned to adhere to timeliness of report generation.

7. Cost benefit : Data processing provides a great advantages in handling, processing and recordingdata in comparison to manual processing.

8. Data storage : A large volume of data can be conveniently stored in computer media.

9. Effective decision making : Because of timely and accurate information to Management, decisionmaking becomes more effective.

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10. Intangible benefit : Reports processed by data processing brings great image of the organizationbefore the customers, industry and community .

5.14 Type of Data Processing

5.14.1 Batch Processing :

Batch processing means processing of a batch of data in one step. This is done only because the physicalsystem generates data during a period of time which needs to be processed at one point of time. Forexample, data for payroll like attendance, advance, other non-standard deductions etc are taken intoconsideration for processing periodically i.e. either monthly or weekly. There is a cutoff date for acceptingdata and then they are batched (grouped) to process.

Batch processing has the following features:

i) Data is collected for a period of time (weekly, monthly, quarterly )

ii) The group of data are entered to create a batch file (transaction file)

iii) Check list is generated in order to identify the errors as validation of some of the important fields arenot done doing entry in off-line mode

iv) Transaction file is edited and then sorted in required sequence.

v) With the batch transaction file and other related files, processing is done in one step.

Advantages of batch processing :

i) It is best suited when the application system generates the periodic transactions to be processed likein payroll or Financial Accounting.

ii) Time requirement of computer system is optimised. One computer can be used for many applicationsystems if proper schedule for batch processing is done for different system.

iii) For large volume of transactions, this processing mode is cost-effective.

Disadvantages of batch processing :

i) Time lag : There is considerable time lag between data generation and report generation.

ii) Reporting System : Because of undue delay in report generation, information sometimes becomeineffective for control measures.

5.14.2 Time – Sharing

Time sharing is a processing system by which numbers of jobs are processed simultaneously. This kind ofsystem is adopted to utilise the CPU time more effectively. The cost of CPU is also high. CPU speed ismuch higher than I/O speed which in normal case results in idle CPU time. There are two different kindsof jobs. The one which takes more CPU time is called CPU bound job and other which takes more I/O timeis called I/O bound job. Time sharing is a process of effective sharing of demand of CPU and I/O devicesby the jobs but the scheduling of jobs by the operating system has to be very efficient. CPU bound jobshave to be assigned higher CPU times whereas I/O bound jobs have to be assigned higher I/O time duringprocessing. CPU time is divided into small time slices and when one job is allocated some CPU time slices,other job is allocated the I/O time. When the second job finishes its I/O operation, the first operation is

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withdrawn from CPU work, either allocated I/O time or kept suspended for some time so that second jobgets opportunity to finish CPU requirement. This is basically a technological development in evolvingefficient scheduling of jobs and corresponding memory management needed for the jobs to be processed.

Advantage of Time-sharing

i) Effective utilisation of CPU time by reducing the idle time

ii) System of processing of multiple jobs simultaneously

iii) High response time

iv) A tool for real time processing

Disadvantage of Time-sharing

i) Problem of scheduling

ii) Problem of memory management

iii) Problem of data integrity and security

5.14.3 On-line Processing

On-line processing means processing of data in an environment where the terminals are attached to theCentral Computer ( i.e. terminals are on-line) so that the transactions are taken care of processinginstantaneously. In on-line environment, the transactions may be processed finally afterwards in subsequentstage for generation of report but the validation of data is being done concurrently. For example, in Bankscorresponding to each transaction of deposit or withdrawal the balance in the accounts is getting updatedimmediately.

5.14.4 Real – Time Processing: Real time-processing provide the facilities for immediate processing oftransactions, updation of relevant master files or databases and fast response as far as output is concerned.Under this kind environment of processing, the output always shows the real-time position of an eventand there is practically no gap in time frame between the incidence of an event and its impact in informationproduction. For example, in Railways Reservation System the database in central location is getting updatedinstantaneously once the input for a reservation or cancellation is entered from any terminal under therailways reservation system and corresponding ticket is printed. Thus, the processing in RailwaysReservation System is real-time processing. Similar is the case of Air-line Reservation system.

A real-time time system is necessarily a on-line system but the reverse is not true. This is because that inthe on-line system, immediate processing is not mandatory and it is different from real-time processingwhereas in real-time processing on-line system is a must otherwise the question of real-time processingdoes not arise.

5.14.5 Multi-programming: Multi-programming is processing a number of programs simultaneously byusing the technique of time-sharing. At a given point of time, actually one job is executed and other jobsare allowed to do input/output activities. This is done for effective utilisation of CPU time.

5.14.6 Multiprocessing: Multi- processing involves simultaneous processing of more than one program byequal numbers of processors. It is different from the multiprogramming. In multiprogramming, one CPUprocesses a number of programs by time-sharing technique whereas in multiprocessing environment, thecomputer system is a multiprocessor one i.e. there are a number of CPU’s in the system. When a job is put

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to the system for processing, it is allocated to a particular CPU with simple algorithm of scheduling and inthe same way a number of jobs can be processed in the system. The sophistication required is in memorymanagement. Large memory is required and memory partitioning for efficient data channelisation isrequired. With the reducing cost of micro-processors and job volume is too large, this system works veryefficiently with less hazards. The cost of the main system is high.

5.14.7 Distributed Data Processing: Distributed data processing is essentially providing facilities to multipleusers at multiple locations through network system under a multi-programming or multi-processingenvironment. This kind of system is adopted according to information requirement at different controllingpoints where transactions out of physical operations at individual points are fed into the system throughthe terminals. For example, this kind of processing is most suitable for an organization having sales officeat multiple places and there is a requirement of controlling the operations of all centers from the CorporateOffice. In such situation a large computer system at Corporate Office with many small processors at differentregional offices and terminals at sales offices dispersed at different locations under communication networksystem can be termed as distributed data processing. In this system, data fed from different terminals willbe processed instantaneously and feedback information is disseminated to all.

Advantages :

i) Reduction in delay in data processing

ii) Effective flow of information

iii) Good feedback information

iv) Support to efficient decision making process

v) Ability to share the data

vi) Direct user interaction facility

Disadvantages

i) Technical problem in connecting dissimilar machines

ii) Problem of data integrity and security

iii) Need of sophisticated communication support

iv) Maintenance problem may defeat the purpose

5.15 Information System Architecture

5.15.1 Centralized : In centralised data processing environment, data files are created on-line or off-line atdifferent locations and processing takes place in a centrally located computer at the head office of thefunctional area. The basic objective of centralised processing are :

• Better control over data and processing

• Sharing of information from large data source

• Lower software and hardware

• Less hazards in operation

• Controlled security of data / information

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5.15.2 Distributed Data Processing

In distributed environment, data feeding takes place at multiple locations simultaneously and processingis done according to the established procedure, normally instantaneously to reflect the business activitiesin the system on-line to make it truly real-time. For Distributed data processing, communication throughinternet is a must. The objectives of this type processing environment are :

• Cheaper hardware

• User-friendly hardware and software

• Instantaneous processing of information from different remote locations

• Privacy of data in stand alone mode

• Personal configuration of PC hardware and software

5.15.3 Client –Server architecture

In Client-server processing environment, there is a main machine (may be a main frame) which is calledserver ( host) and which is connected with several terminals (clients) at different locations for the use byusers. The server software accepts data fed from clients and returns the results to the clients. Networkmechanism provides access permission to multiple users and allow sharing of peripheral devices. Themost important point is taking care of transactions at multiple points on-line and facility of instant updationof database and facilitating the fast dissemination of information to clients users.

In Client Server architecture, the following are the main features :

• It is network based architecture

• Supported by good communication system

• Users are well dispersed

• GUI based operating system

• DBMS software is used

• Open–database connectivity driver (ODBC) and Application Programming Interfaces (APIs)

Advantages of Clients /Server Technology

• Permission of access to system to multiple users simultaneously

• GUI based operating system and GUI at client ends make operation user friendly

• Server and client may have different computer platforms – a flexible option

• Network capability provides good communication facility

• Network facility leads to effective utilisation of computer peripherals from number of clients on sharingbasis

• Flexibility of up-gradation of system both at server end or client ends without much problem

• Management control improves through better information network

• It increases productivity of manpower in information system

It gives high and long term cost–benefit

Communication

STUDY NOTE - 6

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6.1 Communication

Communication through Computers is transmission of signals generated out of data, massage, voice etc.These signals are in coded and decoded to make it meaningful between sender and the receiver. Thisunderstanding has lead to some rules which are called as protocol. The components involved incommunication are :

• Hardware – all types of computer

• Software – that controls the transmission

• Communication Media – media through which electronic signal are transmitted

• Communication Network – link among the computers

• Protocol – Rules for transmission

The development of microcomputer has brought significant changes in communication technology. Thelow cost and flexibility of microcomputer has wide acceptability in communication system. TheCommunication software has been developed to have properly managed communication system withadequate security and integrity of data.

Methods of transmission may be parallel or serial.

Parallel Transmission uses different wires for transmission in different direction whereas serial transmissionuses single wire to transmit each data bit continuously and serially.

Parallel Transmission Serial Transmission

Use of different wires Use of single wires

Hifg speed Slow

Expensive Low cost

Used for high volume data transmission Unsed for transmission between a terminaland a computer

Codes are used to represent characters. Two main codes which are widely used in the world ofcommunication are :

1. American Standard Code for Information Interchange -ASCII – 7 bit code represents 128 characterswhereas ASCII –8 bit code represents 256 characters.

2. Extended Binary-Coded Decimal Interchange Code - EBCDIC – 8 bit code developed by IBM represents256 characters.

6.2 Protocol

When two data communications equipments are connected together, they follow certain rules andconventions while transferring massages between them. Protocols are these procedure and rules for inter-computer communication. Protocols are software that perform a variety of actions necessary for datatransmission between the computers.

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In earlier days, there was not a standard set of protocol which posed a great problem in communicationamong different makes of hardware. The need for development of a standard protocol was felt. IBMdeveloped System Network Architecture ( SNA) which was for communication from mainframe to terminal.United States Department Of Defence (DOD) developed a model called Transmission Control Protocol/Internet Protocol ( TCP/IP) which has become a very standard protocol for network activities.

Protocol determines the following :• Type of error checking

• Data Compression method

• Recognition of massage of sending device that it has finished sending a massage

• Recognition of massage of receiving device that it has received a massage

Some of the functions of communication protocol :

1. Control on information transfer

2. Specification on structure and format of data

3. Recovery steps in case of error

4. Re-transmission

6.3 Modes of data transmission over a channel

1. Simplex - uni-directional where feedback information is not required use simplex communication.Example - Radio, Television

2. Half-duplex – both directional but either sending or receiving at a time. Example - Internet surfing

3. Full-duplex - data can travel both directions at a time. Example -Telephone system

Asynchronous communication is transmission between sender and receiver having independent clocksi.e. any transmission between devices of dissimilar speed like connectivity between computer and printer.

Synchronous communication is Transmission having synchronization between the clocks of sender andreceiver. For example, communication between computer and telephony network.

Characteristics of asynchronous Characteristics of synchronousCommunication Communication

• Simple interface • Complex interface

• Inexpensive • High speed data transfer

• Limited speed – less than 64 kbps

• Requires start and stop bits

• Parity bits often used to validatecorrect reception

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Terminal : Strictly speaking, terminal is referred to device having a monitor and a key board and connectedto a server. If the terminal is having CPU, it is called intelligent terminal. In fact, terminals are the clients ofthe server.

Modem : Modem coverts the digital signal to analog signal and they are carried to the circuit designed foranalog traffic. Modems are used in pairs. At receiving point, it receives the analog signals to convert theminto digital ones which are sent to the computer. A typical modem speed range from 14000 to 56000 bps.

Multiplexer : A multiplexer is a electronic device which carries multiple signals on a single carrier at atime. Multiplexer combines transmission from several computers, transmit the same simultaneously andat the end of the channel with another multiplexer separates them. Use of multiplexer has made thecommunication very cost efficient.

Node : A device on a network. Example – Client, server, hub, modem, gateway, etc. In network, nodes areof two types – end nodes and intermediates. For example, in Telephone network the handsets are endnotes whereas exchange switches are immediate nodes. In computer network, the computers are end nodes.The end nodes do not normally take part in deciding the route of data transmission rather it is decided bythe intermediate nodes.

Route : It is the path between to nodes.

Communication Media : Network media linking the nodes. Example : Network Interface Card ( NIC) orCable or modem, hub, bridges, routers etc.

Repeater : Repeater is used to regenerate the signals to their original strength. While traveling long distance,signals become weak. Repeaters take care of this problem to ensure correct and safe transmission.

Bridge : It is a device within the network to smoothen the transmission by taking care of the load of networkby way of a mechanism of segmenting the network and directing the packets accordingly.

Router : It a inter-network device which is used to extend or segment network to connect nodes across thenetwork. Router decides the path of the packet. Routers are hardware and independent of topology ofnetwork.

Brouter – It is network device having capability of both bridge and router.

Bandwidth : It is the rate of data transmitted unit of time in a communication media. Example – kilobitsper sec ( kbps) or megabits per sec (mbps).

Circuit switching : Circuit dedicated for transmission. Example – Telephone Exchange.

Packet switching : Under it the massage is split into three parts. Header contains information on data inthe packet. Body is the actual data. Footer holds information of error, if any.

One of the fundamental concept in data communication is to understand the difference between analogand digital signal. An analog signal is one which is continuous whereas digital signal may take on discretevalues which are in binary form. Frequency in analog signal is measured in Herz ( Hz).

Transmission of analog signal over a distance requires use of signal amplifier which boost the power ofsignal. Digital Signal transmission uses devices known as repeater. Digital transmission of data is lessprone to error than analog transmission. Other advantages of digital transmission of data :

1. Cost : Due to development of microprocessor and VSLI technology, cost of digital devices are lessthan analog devices.

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2. Data integrity : Data integrity is higher in digital transmission

3. Multiplexing analog signals require expensive communication devices whereas multiplexing of severaldigital signals on communication channel can be done with low cost machine.

Mode of Communication

Different transmission media are given below :

Telephone line ( twisted pair) : The telephone lines are insulated conductors twisted together for the purposeof minimizing noise from environment and referred as twisted pair. It is cheap and widely available forinstallation. It radiates good amount of energy and susceptible to outside noise.

Co-axial Cable : It consists of a central conductor surrounded by insulating inertia. Co-axial cable has arelatively high frequency response due to its low capacitance and has good noise immunity. It is use inunder-sea telephone lines. Its high cost may prove to be a disadvantage in its use in certain installation.Data transmission over a co-axial cable is divided into two type:

Baseband – for analog transmission

Broadband – for digital transmission

Microwave : Microwave transmission consists of high frequency radio transmission. It is generally usedfor high volume data transmission and point-to-point communication. It is used in wide brandcommunication system and is quite common in telephone system. Television system also utilizes themicrowave transmission. Microwave towers can not be placed more than 30 miles apart. To minimize line-of-sight problem, microwave antennas are used.

Fibre optics : Fibre-optic technology is used to transmit the signal using glass fibre. It radiates light ratherthan electricity. It is made of glass and plastic and the cladding part has a different refractive index fromthe core which minimizes the light loss through the sides of the cable and promotes internal reflectionsdown the length of the core. It is very good transmission media in terms of speed and capacity and lesshazards. Fibre-optic cables has made great contribution in reduction of communication media in terms ofsize and weight but enhanced the speed of communication. A single glass fibre can carry more than 50,000telephone calls simultaneously compared to 5,500 calls on a standard coaxial cable line. Speed of opticalfibre in laboratory is six trillion bits per sec.

Satellite Communication : It has some unique features which are given below :

• It has very high capacity

• Transmission is cost independent

• It has a capacity to broadcast transmission over a wide geographical area.

Cellular Radio Technology : A Radio transceiver and a computerized cell–site controller controls all cell-site functions. All cell-sites are connected to a mobile telephone switching office that provides the connectionfrom cellular system to a wired telephone network. As a user travels out from one cell serving area toanother, switching office transfer the calls from one cell to another.

Communication for Information System

The range of communication media available is wide. The Information system select the propercommunication system so that collection of information is efficient and cost effective and according to theneed of the management for business decision making.

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Computer Network

With the revolution in the internet technology and its growing popularity has brought the word ‘network’‘ in common use. A network may be defined as a set of interconnected operation. According to webstar’Dictionary it is defined as “ The act or process of informally sharing information and support, especiallyamong members of a professional group.

In today’s revolution in computer technology, a very significant role is being played by the networkingfacilities among the computers. The concept of distributed data processing is gaining momentum in theworld of business information system. The prospect of linking business units of purchase, production andsales channels through network has been envisaged by the scholars of business information system. Theold model of a big computer serving the need of all functional areas has become outdated. The biggestadvantage the network is sharing information and resources by multi-users.

Operating environment in business world has changed. Railways Reservation System or Airlines ReservationSystem has changed the outlook in business world in India. A Bank is giving us balance immediately onthe terminal, passbook is updated on the printer within few seconds. These all are the boons from Computernetworking system.

Computer network is a system by which computers at different locations can be linked to each other so asto make them enable to communicate among themselves. The ability to communicate is a facility. Theadvantage factor lies in sharing the resources and avoiding duplication of work and facilities of compilationof information from various locations efficiently. The objectives of Computer network are:

• Sharing resource like databases, software, equipments etc

• Saving Cost of hardware and software by sharing them effectively with client- sever technology

• Highly effective information system in a distributed data processing environment

• High reliability in information transmission

Network Administrator – He is responsible for maintenance of a network.

Network Software :

Network Operating system is a software that controls the hardware devices, software, communicationmedia and channels. Two popular network operating systems are Novel Network and Windows NT Server.

Novel Network : The most popular software in network is Novel Network. It is based on client-servermodel. Under this system, a powerful PC should be a server.

Windows NT Server : It is Microsoft’s advanced operating system for networking environment. It has abuilt-in security features. Some features of Windows NT are :

• Data are transferred in 32 bit blocks

• Windows NT gets access to full memory of the system

• Mutli-threading Operating system thereby allowing more than one process to execute at the sametime.

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Network Architecture

The term network architecture refers to either hardware, software or a combination of both hardware andsoftware. The architecture determines the how the network will function. Network architecture may bebroadly classified as either peer-to-peer or client /server architecture.

Information system architecture

Peer-to-peer architecture

Under this type of architecture, each workstation has equivalent capabilities and responsibilities. There isno computer at the centre connecting others through it. It is simpler and less expensive but under heavyload condition their performance is good. Individual resources such as disk drives, CD-ROM drives, printersbecome shared resources which can be accessed by any node. The benefits of peer-to-peer architecture are:

• No Network Administrator is needed

• Network is simple and easy to maintain

• Each computer can provide back up copies of its files to others for security.

• Cost of network is low.

Centralized Architecture

Under this kind of architecture, a single central processor-mainframe or mini computer is located at acentral location and the processing is being done there. Workstations are placed at different locations forfeeding transaction data from there itself. These workstations can avail the feedback information afterprocessing at central computer. Client/ server architecture which is widely being used to-day follows thesame concept of information management.

Client/Server Architecture

Under this type of architecture, at the centre there is a powerful computer dedicated to managing the diskdrives, printers and network traffic management. Less powerful computers (workstations) are clients whichdepend on the server for their access to ant data or resources from the server.

Server - The server is the main Computer in a network environment holding data and other resources (software, hardware peripherals etc.) in a centralized way for enabling the clients to share them. It is basicallyhigh performing computer taking care of network management, security based access control etc.

Client - Client in a networking environment is a workstation through which the users have access to thedata and resources of Server. Server provides the services of information sharing to the clients.

Open systems and Enterprise Networking

Open system means seamless connection of any computing devices, regardless of size and operating system.Hardware connected may range from PC laptop to mainframe. This system represents great flexibility incommunication architecture. This kind of network is a dream goal in technological development. Thiskind of systems are very limited in numbers. The Open system needs connectivity across variouscomponents. Connectivity means ability to exchange application and data from one system to the other. Inmany enterprise now several LANs are seen to be connected with a WAN. This kind of network can becalled an enterprise network.

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Network Topology

It refers to the organization of a network. Nodes in a network are the data processing equipments. Channelsare the connections among the nodes. These channels may be physical or logical. Physical connections willhave continuity whereas logical connections will not.

Star : In Star topology, terminals (end nodes) are connected to a central node or hub node. The features ofthis kind of topology are :

• performance of this kind of topology depends much dep[ends on the processing power of the centralhub

• Terminals have very little active role in network processing and less processing overhead

• Failure of central hub makes the whole system down

• It has the flexibility to accommodate any additional terminals.

Bus : In bus topology, a single network cable runs within a specified area and the terminals (end nodes) areconnected to it. The following are the features :

• Bus network is usually limited to a distance

• There is a need of a routing protocol to manage all the terminals which share the common bus.

• Flexibility provision is there to add any terminal.

Ring : In ring topology, one node is connected to the next and next to the next and last to the first makinga shape of a ring. Features :

• There is a direct point-to-point connection between two neighbouring nodes

• Transmission between the two placed apart has to be done through the intermediate nodes, causingextra overhead of network processing for the others.

• It has also the flexibility of addition of terminals.

Tree : Tree topology is, in fact, a combination of BUS topologies. Several bus LANs are joined to a main busthrough repeaters and bridges to form a structure of tree. Small BUS based LANs are like branches. Featuresare :

• Each node is identified by its unique address.

• It needs routing protocol to ensure correct data transmission to a target node

• The structure is a complex one but provides unique feature to integrate several existing LANs indifferent functional areas.

Local Area Network (LAN)

This kind of network is an inter-connection of computing equipments over a localized area. The geographicalboundary will be within 5 km. and data transfer rate is as high as 1 mbps to 30 mbps. LAN generally usesan assorted topology.

Salient features of LAN :

• Computing equipment are spread over small geographical area.

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• Communication channels between the machines are private

• Server is powerful microcomputer or minicomputer or mainframe.

• LAN file server is a repository of variety of software and data file for the network

• Relatively high capacity communication channels are used

• More reliable in communication

• Cost of interfacing is usually low

• Each device in LAN can work independent of network.

Selecting a LAN

There are several factors that must be taken into considerations for selection of LAN configuration and thecorresponding control features. The following are the important factors :

• Work load – allowable delay under tight condition and controlled delay in heavy load condition

• A baseband BUS topology is simpler to install and maintain. Bus network becomes more reliable withthe use of passive taps. Down time is less.

• In a ring topology, the transmitted signal is regenerated at each node thereby minimizes thetransmission error and enhances the safe distance coverage area. The acceptability of optical fibretechnology in ring network provides high performance possibility

PC bound LAN

One high performance PC is put as file server. Novel Netware software can run on wide varieties ofhardware. File server allows the access to all files on the local disk or shared disk storage. The main featuresof such LAN are :

• Security features allows the network supervisor to grant user authorization and access right

• Files may be sharable, non-sharable, read only, with read/write permission

Wide Area Network (WAN)

This kind of network is an inter-connection of computing equipments over a wide geographical area. Ituses low capacity data transmission technology (nearly 1 mbps).

Salient features of WAN :.

• Computing equipment are spread over wide geographical area.

• Communication channels between the machines are from third party like telephone company, satellitechannels etc.

• Relatively low capacity communication channels are used.

• Reliability in communication is not fully guaranteed.

Hardware Connection

1. Each node is connected to main physical medium through a transceiver unit, also known as mediumaccess control (MAC). (max distance of 50 meter)

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2. Transceiver is attached to the cable with two clamps

3. Ethernet cable is extended with repeaters to improve performance

4. An Ethernet host interface connects the transceiver and controls the operation of transceiver.

5. Transceiver unit is connected to the host unit with five sets of wires:

- carrying power to transceiver

- sending data

- receiving data

- allowing signal a collision to the next interface

- for host interface to initiate isolation of transmit path from the cable

6. The interface accepts the basic data transfer instructions from the computer, controls the transceiverto carry them out, interrupt when the task is over.

Net work Management

The network management involves management of the following :

• Optimal use of resources

• Monitor the configuration of hardware and software continuously

• Right monitoring the performance of network

• Preventive maintenance and fault management

• Security management

Security Management in Network

Security of information is the most important issue in communication network. In the age of Internettechnology, security management in network has gained a serious dimension and there has been gooddevelopment in this area.

This is best managed by controlling the access to the system. Access to the system is controlled by passwordissued to different categories of authorization level of the users.

Firewall

Firewalls offer an effective system to protect access by unauthorized user from outside. The main featureof firewall is packet-filtering router so that vital information does not pass to any unauthorized intruder,even if he manages get access to the network system. It is a system of security in the network with the helpof hardware and software. A software checks all incoming and outgoing internet traffics. The firewallroutes the massages to a safe area to avoid any danger in the in forward transmission of massages. Thescreening by firewall software may delay the transmission process but ensures proper security.

Limitations of Firewall

• Passing on information by internal employees through internet can not be checked

• Firewall can not protect the system from virus

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Cryptography

Cryptography is a technique to encrypt and decrypting massages for maintaining confidentiality in theinformation between sender and receiver.Encryption

Encryption is a process of converting a text into a scrambled form by the use of some mathematical function.Decryption is the reverse process to convert the scrambled form of text into readable text. Generally, theretwo functions – one is used for encryption and the other one is for decryption .Internet

Technically internet represents international network, that means a network connecting the whole world.Internet can also be said as network of networks. These interconnected network exchange informationseamlessly. The power of internet rests on open architecture. Internet has grown to thousands of regionalnetworks connecting millions of users. They are connected with high-speed long-distance back-bone networkwith standard protocol.The biggest achievement of internet technology is that the contents of file is directly displayed instead ofhis downloading the file onto his computers.Internet is a great invention of this age. It has changed the map of communication. The use of internet innormal business communication has gone beyond all estimates and changed the total scenario in the worldof communication. As the world is being developed into a global village, integrated voice and datatechnology is being viewed as a vehicle that will bring about the transformation. Internet is an undeniableexample of technology enabling commerce.The usage of internet has increased boundlessly over the years. In 1998, the users were less than 2% ofworld population. In 2001, it was estimated that approximately 500 million people were internet users.The internet is the massive electronic communication network among the businesses, consumers,government agencies and common people over the world.

Internet usage in 2001

Country As percentage No. of usersof population (millions)

United States 49 134.6Sweden 49 4.4South Korea 41 19.0Australia 40 7.6Canada 38 11.4Netherland 35 5.5Taiwan 32 7.0United Kingdom 28 16.8Japan 27 33.9Germany 24 19.9Italy 22 12.5France 15 9.0Spain 14 5.6Russia 5 7.5China 2 22.5

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Evolution of Internet Technology

1. In1950, the Computer scientists developed technologies that the computers can be linked to eachother and share data, known as networking which opened a new vista in computer data processing.

2. In 1969, US Government established a network among four computers for its defense research andthe network was named as ARPANET (Advanced Research Project Agency Network).

3. In 1970, the network was extended to academic institutions. In 1983, ARPANET was transferred toNational Science Foundation (NSF) exclusive for research and education and a separate networknamed MILNET was put on operation for Defense Research.

4. The APRANET developers felt the need of a mechanism for ‘people-to-people’ communication system.The outcome of which came in 1972 in the form of Electronic Mail (E-mail).

5. E-mail has played a pivotal role in the development of internet protocols and internet engineering

6. The important software TELnet & FTP were developed. Telnet is for having access to host computerirrespective of location and FTP ( File Tranfer Protocol) for transferring files between two computers.

7. TCP/IP (Transmission Control Protocol/Internet Protocol) was developed as a universally acceptedstandard protocol for internet communication.

8. NSF (National Science Foundation) developed a NSFNET connecting five super computers usingTCP/IP technology for the use of universities and research community in USA. Subsequently, NSFencouraged Canadian and European Universities to join NSFNET.

9. In 1984, DSN (Domain Name System) was developed to facilitate the users to navigate with the helpof Domain name. The removed the problem of remembering the internet protocol address of tendigits (IP numbers).

10. By 1985, internet became an established technology.

Function of Internet

Internet has brought revolutionary change in communication technology throughout the world. The futureof this development is beyond the normal imagination. Functions of internet, as on today’s technology, aregiven below :

• Open communication throughout the world through worldwide web services

• Interpersonal communication through e-mail

Internet Connection

For an access to an internet, following facilities are required in the computer :

1. An Account with internet Service provider – Internet address

2. A telephone connection / Cable connection

3. A modem ( Internal or external)

4. An internet browser software ( Internet Explorer/ Netscape Navigator etc.)

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Factors for high popularity of internet :

• Instant communication facilities has resulted phenomenal growth in business efficiency

• Phenomenal reduction in communication cost

• The cost of hardware and technology is within the reach of small entrepreneurs

• Display of information about an organization, its activity and product through websites has becomea common mode of advertisement

• Accelerating dissemination of knowledge and information

• Prospect of E-commerce

• The commercial success of Business Process Outsourcing

• A facilitating media for recruitment etc.

Problems in Internet

While application of internet has opened up marvelous communication facilities in business environment,research and development and at personal level, the problems associated with it are as follows :

1. Security Problem – Internet hackers reveals the password of an user and get unauthorized access tohis confidential and sensitive information causing serious damage.

2. Legal Problem – Question on legal validity of documents transferred through internet communicationis a serious one. When communication is beyond formal one where contractual obligation and otherlegal issues are involved, the problem may pose to be of serious dimension. For example, e-Commerce.

3. Technological Problem – Technological problems are of two kinds. Lack of standardization incommunication mode in terms of file type and support software creates problems in effectivecommunication. The other problem is relating to excessive burden in communication channel duringpeak business hours. In busy hours, the slow speed of internet causes loss of huge productive timeand energy.

4. Spamming Problem – Spamming means sending unwanted advertisement and marketing informationand junk mails to the internet users. During internet operation this causes delay and load to internet.

Internet Service Provider ( ISP)

Commercial internet service providers (ISPs) extends the facilities to have access to various internetapplications and resources for both companies and individuals on payment based on usage. These Serviceproviders are National Service Provider or its agent. They provide very high speed connections to theinternet up to 45 Mbps which is called leased line. Users simply connect the national service provider’snearest switching facility. The ISP then transport the internet traffic across its own network to a routerwith the connection with the other networks in the world.

Leased Line

Leased line is a high speed digital line connection given to the internet users on payment of fees to internetservice provider ( ISP).

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Internet Address

Internet address is assigned to an user. It is for systematic identification of user. It has three parts :

1. IP Address – It is an unique for a particular machine on a particular network. IP address consists offour sections separated by periods for example 201.204.54.16. IP addressing scheme is agreed by allinternet users. IP address is provided by the Internet Service Provider. IP address has the followingcharacteristics :

- It is globally unique

- No two machine can have same IP address.

2. Domain Name – Domain name contains two or more components separated by periods. For exampleibm.com , icwai.org , icai.org

The last component stands for the category of domain name as given below :

com – Commercial company

org – organization ( non-profit making)

edu – educational institution

net – organization directly involved in internet operation.

gov – government organization etc.

3. URL – Uniform Resource Locator

Unform Resource Locator for identification of a particular internet resource.

Generally it has the following structure

Protocol ://server-name.domain-name/directory/filename

For Example http://yahoo.com/apk/index.html

Internet Protocol

There are various internet protocols. Very commonly used protocols are :

1. TCP/IP – Transmission Control Protocol /Internet Protocol

2. HTTP - Hyper Text Transfer Protocol

3. FTP – File Transfer Protocol

4. Telnet – remote login

5. Gopher

6. WAIS – Wide Area Information Service

TCP/IP

TCP/IP is a family of protocol. TCP/IP is the underlying protocol for routing packets on the internet andTCP/IP based network. TCP/IP has two major components i.e. TCP and IP.

Function of TCP (Transmission Control Programme) :

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• Breaks the data into packets that the network can handle efficiently

• Verifies whether all packets have arrived their destination

• Reassembles the data

Functions of IP (Information processing) :

• Envelopes and addresses of data

• Enables the network to read the envelope and forward the data to its destination

• Defines how much data can be fit in a single envelope ( a packet).

Hyper Text Transfer Protocol (HTTP)

The protocol used by www is the Hyper Text Transfer Protocol (HTTP). Hypertext is a text that is speciallycoded using a standard coding system called Hypertext Markup Language (HTML). These links can bethrough text or graphics. HTTP works on client/server principle. The user tries to link the server wherethe resource resides. This is why the address on world wide web begins with http:// . These addresses arecalled Uniform Resource Locators (URLs).

Hyper Text Markup Language ( HTML)

HTML is a kind of language which helps to describe text and graphics by the use of some standard sets ofwell defined tags used for varieties of purposes. The some related texts and graphics taken together toform a page. Pages created with the help of HTML was static in nature in the sense that no informationfrom server could be incorporated in the pages. Now, Dynamic HTML ( DHTML) and Extensible MarkupLanguage (XML) are being widely used. Technology like CGI-Perl, Active Server Page (ASP) have beenevolved to do processing and update database at server end.

FTP

File Transfer Protocol is used for file transfer one computer to another. It works in client-server technology.A client makes request to have an access to information. The FTP client program searches the file, locate itand transfer the file to servers called FTP servers. These servers act as interface and initiate the transferprocess. Steps involved are

• Connection with the FPT server

• Navigate the file structure

• Transfer the file

Telnet

It is a protocol to connect one computer to another. The user which initiates the connection is referred aslocal computer and the other one which accepts the connection is referred as remote computer. The remotecomputer may be physically located in the next room, in the same town or anywhere in the world. Telnetalso works in client-server technology.

Gopher

It is distributed information system that presents information to the users with the help of hierarchicalmenu. Interactions between many gophers on internet allows users to search, retrieve or display informationon different host computers. Information are stored in many computers called Gopher server can beaccessible by a Gopher client software.

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WAIS

It stands for Wide Area Information Service. It is basically an internet search tool which can search morethan one database at a time thus making searching operation fast.

World Wide Web (www)

It is developed by Tim Berners Lee of MIT, USA. The servers are interconnected like a web and network isworld wide. These servers are called web servers. Each distinct file in the web is called web page. A set ofrelated web pages is called web site. The first page of website is called home page. Earlier days, web pagesused to contain only text files. With the advancement of technology, web-pages now contain text, picture,sound and animation. It is also referred as W3.

WWW with HTTP provides the following services :

• Creating links between users

• Incorporating references to picture, graphics etc.

• Communicating with other internet protocol

Web-site

A particular user of the internet facilities may develop a web page. This web page contains information ofthe user. This web page has a definite address which is called the web site address of the user. The otherinternet users can have access to the web site through the address.

Home Page

The first page of the web site is called the home page.

Web Browser

Web Browser is a software which helps the user to contact a web server. Functions of a web browser are :

• Receiving the requests from the user

• Sending the request the web server for information

• Receives the information from the web server

• Sending it the user

The browser is the interface between the user and the web server. Browser allows the user to have access toboth text as well as graphical picture. Browsing graphical picture is time consuming.

Chatting on the Web

Chat means allowing the users directly to communicate to each other. Now the chat facilities have becomevery popular. By way of chatting, the people can interact with each other and exchange ideas and views.There are some sites exclusively for chat. Some of the places where on-line chatting is allowed are givenbelow :

http://www.funcity.com/chat

http://www.talkcity.com/chat

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Search Engine

Search Engine are websites maintaining databases of websites and their contents. When an user wants tolocate a particular information, the task is done through a search engine. Searching is done by index for thedatabase. The web search engine on receiving request from an user, searches the relevant informationfrom the database and submits the information for the user. There are many search engines available onthe web. Some popular search engine are:

www.goggle.com, www.rediffmail.com etc.

Registration on a search engine is free. Search engine is viewed by millions of people every day. They havebecome a strong media for advertisement. Search Engine encourages viewers by providing free service.More viewers means better media for advertisement. Their source of earning is through advertisement. Agreat marketing idea has popularized the use of free internet service.

Managing Website

At the time of designing the website and maintaining the same the following points are to be taken care tomake website more effective in dissemination of information and helping the users in efficient operation :

• Modules should be designed according to the objective of the website

• Web pages should be designed efficiently in terms of information set and cost

• Linking of pages should done scientifically to help prompt downloading

• Website should be updated with the changes in information regularly

• Proper authorization must be there for website management to avoid security problem

On-line Information Service on Web

1. Search : An user can search a particular information with the help of search engine. For example,through search someone can get the names of websites which contains the relevant information onthe subject of his interest.

2. Browsing : Browsing means searching detailed information on a particular subject from differentwebsites.

3. Surfing : Surfing means browsing with any pre-determined search material. It is only browsing overdifferent websites to have any information of his interest.

Electronic Mail

Now, the use of e-mail has become so popular that it needs no explanation to establish its credentials. It ispossible to communicate worldwide over within few seconds. This has revolutionized the world ofcommunication with its magic ability. Any document of file, personal or business-related can be sent withmassages as attachment. This has opened a front to save time, cost and hazards in communication.

What needed are :

1. One Computer

2. Internet Connection

3. User’s name and e-mail address

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e-mail address : Internet connection and e-mail address are provided by Internet Service Provider. e-mailallows the people to send between the computers of the people connected on the internet. e-mail address isthe identification of the sender and receiver. e-mail address is unique world over. The structure of e-mailaddress is

[email protected]

example : [email protected]

Mailbox – Mailbox is used in e-mail to store in-coming message. The receiver is not necessarily be presentwhen the e-mail is delivered to him. All his message will be stored in the mail box.

Mail Server – Mail server basically sends a mail to the e-mail address of the receiver. It does the following:

• It maintains list of e-mail account

• It allows to compose message.

• It sends the message when send button is clicked.

• It arranges the messages of the receiver in the order they are received etc.

Address book – An address book is a place where e-mail addresses of people with whom frequentcommunications are made are kept. New addresses can be added to it or some existing addresses may bedeleted.

File Attachment –The major drawback in e-mail is that the formatted text can not be sent. If a word filecontaining graphics picture or bulleted list etc is sent, only text will be available and all other things will belost. The solution for this is to send the document as attachment file to the e-mail address. It also saves timeand money.

Activities under e-mail

1. Composition : Composition of message preparation may be on-line or off-line. On-line means it donewhen internet is active. In case of big message, the same is prepared off-line and is sent afterwardsaccording to convenience of the sender. Files can be sent along with the message as attachment.

2. Sending / Receiving : Sending or receiving message means transmission of message from sender toreceiver. The sender puts the e-mail address of the receiver and the message is transferred to receiver’se-mail address through the server. Receiver receives it whenever he logs in at his internet addresswith the password. If the receiver’ e-mail address is wrong, e-mail will be bounced back to sender.

3. Reporting : Reporting means giving the sender the status of message sent.

4. Display / Download : Receiver can see the message on the computer or he can download the contentsin a file or on printer.

5. Replying and Forwarding e-mail massages

This is very good option provided by e-mail program. By pressing the button, the option can be availed.The advantages are :

• Address of the person to whom reply is being made, will automatically come

• Subject also be written automatically with prefix ‘Re:’.

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• Original massage can be sent back with proper option for ready reference

• ‘Reply to author’ or ‘Reply to all’ option is exclusively for author or all to which copy was sentrespectively.

6. Deletion – Messages after receiving may be deleted to arrange space in the mailbox.

Effective way of e-mail use

• Massage to be composed off-line to save on-line time

• Address book may be used to save time to pick up the addresses with whom there are frequentcommunications

• Mails may be organized in different folders to keep track

• Check the mail box preferably during non-peak hours so as to get quicker connection

Advantages of e-mail – mail has become very effective mode of communications in business environmentor at personal level. The advantages many which can be summarized as below :

1. Speed – The communication through electronic media is instantaneous from one corner of the globeto the other.

2. Cost – Cost is very low compared to any other mode of communication.

3. Sharing information – Same message can be sent to multiple receivers at one go giving all e-mailaddresses of individual receivers.

4. Documentation of communication – An e-mail keeps date and time which is preserved with themessage. This is useful for the purpose of auditing the communication in case of future complicacy.

5. Powerful media for communication – The multimedia combining text, graphics etc. can be sent asattached files.

Voice Mail

Voice Mail is a system of recording the voice of the speaker (caller) on a recording media and massage ispassed on the appropriate user. This system is used in case of STD calls. It has the advantage of recordingthe massage for future use by the appropriate person. Even the massage can be distributed to differentusers through e-mail after conversion of massage in digital form.

E-Commerce : The possibility of almost frictionless access to information globally has brought the idea ofbusiness transaction through internet which in termed as E-Commerce. IBM has promoted the E-Commerceconcepts and tools through different solutions and developments. The concept of E-commerce is that theconsumers can have access to global market, advantage to choose the best, free option to explore the termsand conditions for procurement of an item from different vendors etc. These can be done sitting at home oroffice without having the painful exercise of visiting market and negotiating terms and conditions. Theinstantaneous gathering of information and providing decision based on pre-determined criteria are to betaken by the system.

To understand how it works, let us see what are the components involved in it :

• Customers

• Suppliers

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• Service Provider

• Channel partner (distributor)

• Regulatory authority

A consumer interacts with the online system through web browser. A consumer first interact then useshyper link to access a shopping mall. A Shopping mall is where a customer first visits for shopping spreeand there may be many pages in the shopping mall. Business process follows the path :

• Consumer selects a store

• Link to the merchant server

• Customer selects an item from the e-shop

• Receipt of payment by supplier through credit card

• Obtain payment authorization

• Physical delivery of the item

Transactions in E-Commerce

• Consumer accesses a shopping mall and select a shop for purchasing certain item.

• Shopping mall serve the access to the merchant for a selected shop.

• Merchant system presence the home page to the consumer.

• Consumer selects the desired goods.

• Consumer interacts with the merchant system and make the payment.

• Merchant system accesses his bank for authorization of the consumer payment.

• Authorization of payment by the bank.

• Merchant system informs the consumer that the payment is accepted and transaction is completed.

• The consumer bank informs the customer of the money transfer.

Advantages of E-Commerce through Internet :

1. Easy access to global information : The Suppliers will put their product technical specifications andother commercial information in the world wide web. This publicity will involve little cost. Customerscan make enquiry and place order through internet. Easy access to internet will give him opportunityto compare terms and conditions for commercial transaction.

2. Effective Advertising Media : The consensus on low cost provision for advertisement in www will bea advantage in the hands of suppliers to open up their products to new customers who prefer internetfor their shopping needs.

3. Free Entry : The powerful search engines are encouraging the users to have access free of cost. Theprospective buyers can take the advantage.

4. Market driven strategy : The suppliers can judge their product response by analysing the visitors tothe website and number of matured order and start their strategic decisions to carry on with thepresent trend or go for luring more buyers with better offer in terms of quality and price.

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Impediments in E-Commerce

The research conducted by many experts on impediments of e-Commerce. Some of them are identifiedand given as follows:

1. Security : When an organization uses the internet to engage in e-commerce, it is likely many of itsinformation are exposed to security risk, fraud and abuse. Out of them the most serious is credit cardinformation.

2. Legal Issues are many like protection against fraud, passing sensitive data to strangers etc.

3. Cost of hardware, software and maintenance

4. Lack of expertise

5. Need of training

6. Uncertainty of market

Issues covered under E-commerce

• Legal issues relevant to e-commerce

• Evolving business practices

• Relevant legal standard

• Proper guidance

• How to create enforceable digital contracts for small goods or services

• Digital transaction will be enforceable and valid as a traditional paper based transaction

• Signature as an evidence

• Time, date-stamping to provide proof of dispatch, receipt, submission etc.

• Authentication procedure

• Legally binding commercial transaction etc.

E-commerce is progressing fast but legal and control provisions have not yet been developed in the formof insuring valid contractual obligations. Thus there is a need to develop legal mechanism. The law hasbeen slow in relation to the progress of technology. Parties to electronic commerce transaction have workedwithin the system of traditional, contract law, making adjustments to fit into the new business practices.

Validity of agreement and enforceable by law

Two requirements of a valid contract :-

i) Mutual ascent / intent by the parties to come into contact

ii) Sufficient definite terms

Here definite terms involve mechanism on actual performance, such as delivery of goods or services,payment for them or return them.

The provision for enforcing a compliance of a failure to perform according the contract is a breach ofcontract. Fundamental objectives of contract law is to protect a party who accepts a promise in a properly

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framed agreement, from injury as result of breach by another party. Law enforces various avenues ofrecourse to enforce his right under the contract law. Injured must proof his injury and damage and theparty who breaches may face several types of liabilities.

Binding of contract

• Protection of system error, programming error, human error.

• Evidence in case of digital agreement

• Invoices and payments related to e-mail must be authenticated

• Security policy defines some of the controls and counter measures that must be provided.

Auditability

In order to resolve, the parties involve will produce evidence to support their cases. The principal securityrequirement here is that the integrity of the audit trend must be assured. Integrity of the electronic data filemay be manipulated without any trace. It will create lot of problem. Necessary security service is essential.Proof of source authentication and integrity of messages are to be preserved.

Virtual Reality

Virtual Reality is a programming language to simulate the model in a such a fashion that it looks like real.This tool is being thought of to be useful in advertising in web-sites for an effective e-commerce business.

Electronic Cash

Electronic cash or e-Cash or Digital Money is to provide the means to transfer money between parties inthe network. Steps involved are :

• Consumer request his bank to transfer money to e-mint to obtain electronic cash.

• Consumer bank transfers money from customer account to e-mint

• E-mint sends the cash to consumer. Consumer sends the electronic cash in a hard drive or a small card

• Consumer selects the goods and transfer the e-cash to merchant

• The merchant provide the goods to the customer

• The merchant sends the electronic cash to his bank

• The bank redeem the money from the e-mint

• e-mint transfers the money to the merchant’s bank account

Benefits of Electronic Cash

• Potential fraud is reduced- when the bank receives the electronic cash, it verifies the serial number, itdeletes the serial no and takes it out of circulation forever. The same number can not be re-used.

• Preference of Merchant – Merchant will prefer an electronic cash, since it guarantees the payment

• Confidentiality of customer – Customer’s confidential information regarding his Bank Account etc. isnot disclosed.

Digital Subscriber Line Technology (DSL technology) : DSL technology is a powerful technology. It willhave three compelling advantages :

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• High speed data transfer rate (128 Kbps to 6Mbps)

• Meeting the need of growing brand-width for application need

• Low implementation cost

Electronic Data Interchange (EDI)

Electronic Data Interchange is a process of exchange of structured business information (documents) amongvarious agencies involved in the process of transaction through electronics. Purchase orders, invoices,remittance advice etc. are the documents which need to flow smoothly in the business process. EDI is theservice provider in the process.

EDI’s job is to receive and transit the documents among the trading partners. Traditional EDI communicationroute is through value added network (VAN), which is a third party service provider that receives, storesand transmit data. EDI’s process comprises three sub-systems :

1. Translation : EDI software converts files from trading partners into EDI standard format, called EDIdocument

2. Transmission : EDI documents are transmitted using mutually agreed communication method

3. Retranslation : When a trading partner receives a transaction, it is retranslated with the help of EDIenabled software into a format which can be used as its own business document format.

EDI documents are developed for the following reasons :

• To ensure smooth transmission of data

• To make data suitable for storage in database

EDI will be necessary requirement for Indian exporters to exchange business information and documents.The US, India’s largest market will make it mandatory

To flow data through EDI for international trading transaction. In India, EDI will start operation in internetbusiness. RBI, Ministry of Commerce, Customs Department, Airlines, Airport, Seaport, DG Foreign Tradeetc has already agreed for implementation of EDI documentation in foreign trade. Major problem has beenthe inter-connectivity of the various organization.

Advantages of EDI

1. Reduced manual entry leading to an increased accuracy

2. Reduced clerical overhead resulting lowering cost

3. Faster delivery of data

4. All business transactions are immediately acknowledged.

Electronic Fund Transfer

Bank to Bank Transaction and commercial transaction through electronic needs confirmation andauthentication. The transactions involved in the process are :

• Buying and selling of foreign exchange

• Inter-bank deposits

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• Payment through credit card

The major issue of security against data leakage and protection mechanism from Merchant fraud.

Secure Electronic Transaction (SET)

Master Card and Visa International have been developed. SET protocols which are :

• Merchant will have no access to credit card information

• SET requires authentication of trade parties prior to commencement of processing

System adopted by SET :

Encryption of both information with secret keys :

i) Financial information : Credit card information which is a private matter between card holder andcard issuing agency

ii) Non-financial information : Details of goods purchased, mailing instructions, cost etc. which areprivate information between trading parties.

Intranet

Intranet is a in-house version of internet. Intranet is a private network. It uses the same mechanism ofcommunication - internet software, TCP/IP protocol and web browsers. But there is difference in terms ofbehavioural environment. Internet operates in open environment whereas intranet operates in controlledenvironment. Typically intranet operates between different offices in an organization. An intranet is aperfect solution for small and disciplined team scattered over a wide geographical area. Companies areincreasingly using intranet to broadcast internal information generated out of transaction processing atdifferent points among themselves and to the controlling branches. Following are basis features of Intranet:

• Based on internet protocol which can be expanded worldwide.

• Based on architecture with flexible expandability

• Low cost compared to internet facilities

• Facilities of handling multimedia data

• Communication facility limited to the organization – better control on data integrity

• Serves the work of information mobilization for MIS

When a corporate organization has its sales office dispersed over a wide area and plans for closer controlover the target market with the help of better service, it needs continuous feed back from market. Theintranet mechanism can help it to achieve the goal with ease and in a more cost-effective way.

Specific Application of Intranet

1. Sales Management

• Production information

• Price list

• Information on sales

• Analysis of local competition etc.

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2. Customer Service

• Conduct training to Customer Support Team

• Determine the status of particular product / Customer

• Latest orders status

• Publish technical support bulletins etc.

3. Accounting Information

• Monitoring company’s stock price

• Publish the public annual report

• Preparing list of debtors etc.

Extranet

An extranet is a business-to-business intranet i.e. inter-organizational information system. It is an extensionof intranet concept linking all intranets of related organizations for mutual benefits and sharing informationamong themselves. Instead of sharing information within the information within one organization, it goesbeyond. Generally it is opted when the two organizations are closely related by way of same group ofentrepreneur house or they are strategic partners for a common corporate objective. In extranet, sameinfrastructure components as the internet, TCP/IP protocol and Web browser are used.

On Extranet, each participating companies have to have a common understanding for a common objective.It may be sharing of information or sharing of software or sharing of technology in the computer media.Extranet is commonly opted among the companies involve in supply chain management.

Benefits of extranet :

• Better corporate communication environment

• Low communication cost

• Faster processing of information flow etc.

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7.1 Data Source

Data source may be internal or external. Internal data are from transaction processing of the organization.External data are those which are collected from outside an organization but relevant to the organizationfor its use. These data may be from a published reports from survey, studies, observations or from internetor from other sources. Example –

• Market information

• Product information

• Technology information etc.

7.2 Data Quality

Quality of data is the most important aspect as far as its use is concerned. The classic expression on theimportance of quality of data for data processing is represented by the word ‘GIGO’ – Garbage In GarbageOut. This word was devised to forewarn the systems people to be all careful for ensuring quality of datafor the success of an information system. The poor quality of data means incorrect, incomplete andambiguous data. A poor quality data may cause great damage in reporting and action initiated by it. Theaim of good quality of data is to provide information which must have these quality:

• Relevance

• Timeliness

• Accuracy

• Completeness

• Consciousness

• Understandability

7.3 Data Storage and Management

The amount of data increases exponentially with time. The requirement of preserving long-term data hasbecome a great burden on an information system. Of course, the storage capacity of computer has becomevery high and the cost of storage has gone down also. With this development in information technology, asystem favours preserving long-terms data for better in-look in the information analysis. The basic advantageof a computer system is the standard organization of data set and ease of retrieval. The factors which guidedata storage are:

• Requirement for analysis and purpose of use

• Selection of data management tools

• Data organization

• Adherence of legal requirements

• Security of data etc.

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7.4 Data Warehousing

The data warehousing is the concept of data integration by way specialized data storage and retrievaltechnique. The core of data warehousing is multi-dimensional databases. The basic objective of datawarehousing is give right kind of expert analysis of data and feed the decision makers with right kind ofinformation for effective decision to run the business efficiently.

The need for data warehousing has come from increased competition and demand for more and accurateinformation for precise decision making. Data warehousing can translate the data in common format andstore them scientifically for faster access. Combining the related data is the crux in the approach of datawarehousing. Thus, database design has a definite key role in making the storage efficient. Providingsensible information is only possible through analysis of related data.

Steps involved:

• Development of hardware and software infrastructure

• Building sound corporate database

• Establishment of communication network

• Managing smooth data flow from multiple operational points

• Building proper check and security measures from misuse

Critical factors involved in data warehousing:

• Heterogeneous data set and their translation in common format

• Organization of database

• Capability of software tool

• Judgement on transactional period for storage of data

Advantages:

• Effective, efficient and convenient data storage

• Elimination of duplication

• Lower storage of cost

• Faster navigation through related information

• Reliable data

• Consistency in reporting

• Better decision making

Strategic use of data warehousing in different business environments:

• Customer Service

• Sales Promotion

• Market analysis of new products

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• Trend Analysis

• Inventory Control

• Sales decisions etc.

7.5 Data Mart

Data mart is the simple form of data warehousing. In other words, it is a scaled-down version of datawarehousing. Data marts of a company are generally created with specific objectives. It may be functionspecific. The advantages of creation of data mart are low cost and less time requirement. Data marts arecreated with a specific focus. For example, data marts may be created for marketing department withcompetitors information only to develop business strategy to improve market share. This kind of approachis meaningful as far as relevant information base is required. Instead of waiting for comprehensive datawarehousing, data mart sometimes provide tremendous services to meet immediate and specificinformation need.

7.6 Data Mining

Data mining name is derived from the similarity of searching out valuable business information from ahuge database and mining of valuable ore after locating it from below a huge mountain or sea. The datamining is the result of long process of research on possible methods of proactive information delivery.Data mining is extraction of data from multiple data sources by way of interactive and analytical softwaretools. In other words, it is the process of finding the relevant information from a large database with thehelp of ‘expert’ software and using the analytical tools for data analysis which turn them ‘knowledge’ forthe decision makers. The decision makers’ job becomes easy as they have less botheration on how theanalytical tools work. They make queries and results based on ‘expert tools’ flow in their hands to makethe process more easy going, sophisticated and dynamic.

Data mining involves intelligent probing on data to make reliable and predictive information from a largedata base. The software tools available for data mining provide the following types of services:

• Neutral computing based on data available to identify potential areas and suggest precautionarymeasures in the business process

• Intelligent agent’s work in retrieving most reliable information

• Support of special expert with necessary IT tools to extract appropriate dataset and use goodmathematical model to make predictive solutions

• Optimization of business process variables with Operation Research techniques and guiding businessdecisions

• Evolving industry standards on many business processes

Advantages of Data Mining:

• It provides finite set of values of variables with other relevant information to understand their trendand behaviour

• It determines the significant and possible variations in the variables

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• It helps in building predictive model for future values of target variables

• It is a guiding force for proper decision making

7.7 Knowledge Management

Knowledge is a mix of information, experience, evaluation. Thus, information system in an organization isthe key resource on which knowledge base depends. The main function of knowledge management incorporate world involves the process development of minds of decision makers through continuous flowof right kind information and their critical evaluation. Steps involved are :

• Creating the information data base and constant updates

• Creating technological infrastructure

• Developing analytical skill

• Continuous feedback system

• Creating culture of sharing – exchange of information

• Capturing knowledge (analytical information and expert opinion)

It is a process of accounting and creating knowledgebase for the organization for facilitating sharing ofknowledge by managers for efficient management throughout the organization. The crux in exercise ofknowledge management is to transform the data warehouse into knowledge base. Knowledge base means:

• Best company profile

• Performance measurement tools

• Best control practices

• Complex business environment parameters

The knowledge management has demonstrated a good account of success in deriving effective businesssolution. This tool is in developing process and its potential is enormous in guiding the model managementin making decisions relating to market, product, technology and so on.

• It is an intellectual capital

• Knowledge base is assimilation of best experience and business process

• It is a collection of effective business solutions history

• Comprehensive corporate data base.

The process of creation of knowledge management involves the great deal of technical tools which are:

• Knowledge identification

• Storage

• Organization of information

• Retrieval system

• Dissemination system

• Effective use

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Knowledge base stands in one place on the organization in a structured manner so that it is accessible bythe users in the organization. The aim of knowledge management is to guide the managers in takingappropriate decisions, prevent managers to do the same mistakes and forecast on business outcome andshowing the path of success.

Sources of Knowledge.

• Information base

• Past experience

• Expertise in specialized field

• Sophisticated analytical ability with mathematical tools

Benefit of Knowledge Management:

• Increased ability to compete

• Richer knowledge stock

• Effective utilization of resources

• Stronger ‘base of ideas’ for innovation

Information requirement for Management Accounting has the following objectives:

• Collection and presentation of information relating to business

• To facilitate control

• To improve the users knowledge

• To add the value of the product or service

• To plan for the future on the basis of the information on customer’s opinion, computer behaviour,and new technology

Social and organizational impact:

• Impact of development on telecommunication

• Interpretation and use of intelligent agents software

• Cultural dimension of IT acceptance

• Management of chain and potential staff reaction

• Growing awareness of remote working

7.7.1 Knowledge Management for Innovation framework:

• Identify business arrears and processes that are suitable for innovation

• Identifying change levers; those tools that can be used to innovate

• Development process vision; statement of purpose for the processes

• Understanding the existing processes

• Designing and prototyping new processes

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7.7.2 Innovation culture:

• Change in the culture of the organization

• It is not sufficient to allow managers to invent new processes by stumbling across opportunities

• They should be encouraged to be continually on the look out for new and better ways of doing things

• The value chain analysis

7.8 Business Process Re-engineering (BPR)

Business Process Re - engineering is a continuous process of rethinking, re - assessment, re - designing andre-valuation of each elements of business process. Re-engineering gives stress on bringing changes toimprove performance. It comes through 6R’s.

Business Process Re-engineering

• Realization for change

• Requirement

• Re - thinking

• Redesign

• Retooling

• Re-evaluation

Success of the BPR lies in correctly understanding the shortcomings in the existing system and risk in thechanges and undertaking the technical feasibility study of the proposed system. BPR is an all embracingexercise to bring changes in business process such as overhauling organizational structure, improvingmanagement system, bringing in new work culture and replacing old business system with a new one.

Success of BPR depends on effective planning on what are the changes to be brought, what will be theirpriority and how to effect the changes. Obviously, it involves a great deal of trial and error exercises. Rightkind of analysis of causes of failure is needed. The knowledge gathered gives the right direction to avoidmistakes and to ultimate success.

Most critical part of implementation of BPR is transformation of workforce i.e. changing human behaviourand attitude. Success index depends on extent of co-operation from the employees of an organization.

Origin and scope of BPR:

• Fundamental rethinking and redesigning of business processes to achieve dramatic improvement incritical measure of performance such as cost, quality, service and speed.

• Process identification – an organization or a department is broken into series of processes.

• Process rationalization – identification of the value addition by each process and elimination of non-essential processes.

• Process redesign – the remaining processes are redesigned so that they work in most efficient way.

• Process reassembly – the re-engineered processes are implemented resulting in efficient performancein tasks, department and the organization as a whole.

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Steps involved in BPR:

1. Begin Organizational Change

• Assess the current state of organization

• Understand the existing process

• Explain the need for change

2. Develop vision for change and success

• Communication campaign for change

• Making employees understand the need for change

• Projections of outcome of change

• Remove the fear psychosis on security of job

3. Devising BPR plans and programmes

• Evaluate the existing business strategies

• Discuss the issues relating to customers’ desire

• Identify new business opportunities

• Select the processes to be re - engineered

• Identify potential barriers against implementation

• Formulate new process performance strategies

4. Change in work practice and culture

• Brainstorm using BPR principles

• Use customer value as focal point

• Explain the new values and culture required

• Establish re-engineering teams, cross functional workgroups

• Allow the workers to make decisions

5. Establish new business system

• Prepare document for new organizational structure

• Draw new work flow diagram

• Create model on new process steps

• Describe new technology specifications

• Redesigning the information flow requirement

6. Perform the transformation process

• Develop a migration strategy

• Map new skill requirements

• Train the employees on new processes

• Reallocate the workforce

• Transform the organization

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7.9 Expert System

According to CIMA, an Expert System is an application software system which is used to store data relevantto a particular subject area and to provide solutions to problems requiring discriminatory judgement basedon that data.

Expert system is a software which derives extraordinary intelligent solution like an expert. The knowledgeof an expert in invaded in the software with solution options for different complex problems situation,particularly, unstructured problem situation. Here the expert knowledge is knowledge of specialized fieldand solutions sets at different problem situations.

For example, knowledge of expert marketing management for experts system in marketing, knowledge oflegal expert for expert system in legal field, expert knowledge of taxation for expert system in taxation.This expert knowledge and history of different unstructured problem solutions are stored in organizedmanner so that the related expert system can use the data base. The expert system is not a simple managementinformation system. Rather it helps in involving solutions in complex problem situation. The componentin expert systems are

• Data management

• Expert knowledge handling tools

• Complex problem situations and framing corresponding solutions sets.

7.10 Artificial Intelligence (AI)

Artificial Intelligence refers to behaviour of computer system which seems to be intelligence of the computeritself. As you know the computer does not have power to act on its own, the intelligent solutions from acomputer system is termed to be Artificial Intelligence. In fact, this phenomena is exhibited by a computersystem out of the expertise of the software which has the capability to interpret the problem situation anduse the knowledge base to evolve intelligent solutions. The artificial intelligence is closely associated withexpert system. It is only the expert system which can exhibit artificial intelligence.

7.11 Decision Making Process

Decision making is an important managerial function of choosing a particular course of action out ofseveral alternatives. Developing decision making system is an integral part of an organization. For a gooddecision making system, the following criteria must be followed:

• Creation of fair, free and sound decision making environment

• Establishment of ground rules for decision making

• Formulation of organizational policy and procedure

• Establishment of proper Information system

• Training to junior managers on judgement skill and art of decision making

Decision making is a process to improve operational efficiency of the organization. The objectives of decisionmaking are:

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• To make effective use of resources

• To have better control measures

• To improve operational efficiency

• To augment profitability

• To face competition

Steps in decision making process

Decision making process follows a scientific system and the steps involved are:

i) Identification of problem and diagnosis of the problem

ii) Collection of relevant Information

iii) Search for alternatives and evaluation of them

iv) Decision structuring

v) Selecting preferred option

vi) Communication to relevant action points

vii) Implementation

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The following factors determine the quality of decision making:

• Analytical ability and experience of the decision maker

• Flow of relevant information

• Tools available for information analysis

• Reconciliation with unavoidable factors

Timely implementation

Programmed and Non-programmed decisions

Programmed decision making refers to those decision making process which are based on some standardset of procedure established by the management and according to scientific principle of management. Incase of programmed decision making, supporting information sets and reports are standard, well definedand well structured. Naturally decision making process is simple and based on some guidelines. Forexample, stores ledger summary and material consumption reports may help in decision making onInventory control.

Non-programmed decision making refers to those decision making process which does not go by any pre-determined set of guidelines. Normally this type of decision making takes place to handle special businesssituations with the help of experience, judgement and vision of the decision maker. In case of non-programmed decision making, information are unstructured and external environmental information is amust along with internal information sets. For example, for decision on business policy many non-standardinformation like technology change, competitors market share etc. is required apart from internalinformation of sales of different products.

Decision hierarchy

Decision making process follows a well established hierarchy. Different levels of management are involvedin different types of decisions. This type of decision depends on the level of responsibility and theircontrolling activities. Decision may be of different kinds. The diagram showing the decision hierarchyexplains lower level management takes decision on operation, whereas middle level on tactical and topmanagement on business strategy.

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Top Management – Top management is concerned with strategic decisions like diversification, technologyacquisition, new market exploration, strategic business alliance, takeover, merger etc.

Middle Level – Middle level management is generally involved in tactical decision making with the help ofperformance analysis, budget variance analysis, devising better productivity mechanism and control etc.

Operational Management – Operational management staff are mainly involved in scheduling the activities,keeping track of progress of day-to-day operations and decisions of well structured problems etc.

Strategic Decision Making

Strategic decision making is concerned with the issues relating to long-term effect on the business growthand prospect of the organization as a whole and involves both internal and external factors. Strategicdecision making process involves the following:

• Assumptions of possible future state of business environment

• Generating alternative strategies

• Enhancing the problem prevention capabilities

• Decisions based on different business scenario

Important Financial decisions:

• Capital structure decisions

• Capital budgeting decisions

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• Working Capital Management

• Asset Management

• Tax Planning

• Profit Planning etc.

Important decision in Marketing Management:

• Product pricing

• Sales incentive

• Sales promotion activities

• Market Research and intelligence

• Product development strategy

• Product pricing

• Customer service etc.

Important decision on Production Management

• Volume of production of different products

• Schedule of production

• Manpower deployment

• Cost control

• Quality control measures etc.

Important decisions on Human Resource Management

• Recruitment policy

• Placement

• Wage negotiations

• Training scheme

• Incentive scheme etc.

Important decisions in Project Management

• New project/modernization project

• Technical consultancy

• Investment planning

• Long-term and short-term project funding

• Equipment procurement etc.

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Tactical Decision Making

Tactical decisions are concerned with issues having short-term effect on the business. Primary aim of thistype of decision is to effective use of resources, remove any imbalance among different factors of productionand improve productivity of factors of production. Often decisions are related to acquisition of resourcesto help strategic decisions.

Operational Decision Making

Operational decisions are related to the effective use of resources acquired out of tactical decisions and aimof these decisions are to have effective control of the use of resources in day-to-day business activities.Decisions are directed to achieve the target as decided by tactical decisions.

7.12 Decision Support System (DSS)

According to CIMA, Decision Support System (DSS) is a collective term to describe the various types ofsoftware, principally modelling or simulation, which provide input into management’s decision makingprocess. The ‘what if?’ analysis produced by spreadsheets is a typical example.

Decision support system is an IT tool based on models for interpretation and analysis of data and presentingthe same for facilitating decision making. An expert system is within the DSS so that unstructured problemsare also taken care.

Decision Support System has the following three characteristics:

i) Support semi-structured and unstructured decision making

ii) Use model for solutions

iii) Flexible to respond the changing need of the decision maker

Structured or Programmed decision making is with the help of using data in a simple related decisionmaking program which is based on technical knowledge on a particular area.

Unstructured or semi-structured decision making process involves a great technical expertise of the allsubjects of all related fields and building model to handle the information to derive solution set.

Decision Support System

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7.12.1 Components of DSS :

A DSS has the following three components:

i) Databases – A DSS must have one or more databases containing all relevant information. The systemwill extract the set of information from database with data-mining tools.

ii) Model base – Model base is the brain of a DSS. For example, mathematical models on Time Seriesanalysis, Linear Programming, Statistical Quality Control will be tools used to solve many financialproblems.

iii) Technical Expertise – A DSS system must have technical expertise in the domain field to understandthe nature of information, nature of problems, appropriateness of model and additional intelligencerequired to handle the unstructured or semi-structured problem situations.

7.13 Flow of Information for supporting decision making

ManagementInformation

System

STUDY NOTE - 8

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Management Information System

8.1 Management Information System

Management Information System is a systematic process of providing relevant information in right time inright format to all levels of users in the organization for effective decision making. MIS is also defined to besystem of collection, processing, retrieving and transmission of data to meet the information requirementof different levels of managers in an organization.

According to CIMA-

MIS is a set of procedures designed to provide managers at different levels in the organization withinformation for decision making, and for control of those parts of the business for which they are responsible.

8.2 Objectives of MIS

• To provide the managers at all levels with timely and accurate information for control of businessactivities

• To highlight the critical factors in the operation of the business for appropriate decision making

• To develop a systematic and regular process of communication within the organization on performancein different functional areas

• To use the tools and techniques available under the system for programmed decision making

• To provide best services to customers

• To gain competitive advantage

• To provide information support for business planning for future

8.3 Basic Features of an MIS

8.3.1 Management Oriented: The basic aim of MIS is to provide necessary information support to theManagement in decision making. Naturally, the design of MIS is framed keeping the consideration ofrequirement of management of the organization. The management plays the most important role to developthe system specification of MIS to ensure the information need of different levels of management.

8.3.2 Timeliness: One of the most important issues involved in the effectiveness of MIS are flow of informationat right time to the user level of management.

8.3.3 Integrated: Disintegrated information from different sub-systems becomes too much clumsy in decisionmaking. Information from different related functional areas is to be integrated to make it meaningful forthe users. In today’s business environment, the different parameters of business process have a complexequation in an overall impact on the outcome. To understand the combined effect, linking them with thehelp of different analytical models is essential.

8.3.4 Simplicity: An MIS System should be as simple as possible so that people at operation and users donot feel any hazards. The success of a system lies in the acceptance by operation staff and users. A goodamount of planning is required on the part of a MIS designer to make it simple keeping in mind the needof information at different levels of management. The deigning skill of database and analytical modelsdecides the quality of MIS.

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8.3.5 Reliability: MIS system should provide most reliable information. For that what is needed the skill indesign and efficiency in programming to develop the system. More important is a thorough check of inputinformation, process flow and output reports on regular and routine basis.

8.3.6 Consistency: The input data and output reports must follow some standard norms so that consistencyis preserved.

8.3.7 Relevance: Only relevant information flow at different levels of management helps in increasing theeffectiveness of MIS. In other words, information of right quantity improves the efficiency of the users inthe use of information for their decision making.

8.3.8 Flexibility: MIS should be flexible enough to take care of changes in the environment in the businesssystem.

8.4 Implementation of MIS

For establishment of MIS in an organization, the following steps are followed :

8.4.1 Analytical study on information requirement: A joint efforts by systems experts and managementexperts is required to understand the exact need of information at different levels of management and howto assimilate them from data flow from different sources. The anticipated change in the need of informationmay be kept in mind while planning the design in order to provide sufficient flexibility in the system.

8.4.2 Determine the sources of information: Once the first step is understood, it is to see how to get therequired information and their sources. If required, data recording system may be changed at differentpoints so that exact data flow is ensured and the same can be done without much hazards. For the sake ofsimplicity of the system reorientation in the physical flow of data has to be done.

8.4.3 Establishment of right kind of data processing environment: The important step involved in MISdesigning is arranging the right kind of tools for processing i.e Computer System and infrastructure interms of software and skilled manpower. The proper scheduling of processing is equally important toensure smooth flow of information.

8.4.4 Selection of software: One of the important factors of success for MIS is quality of software. Softwaremust fulfill the following criteria :

• Compatibility of hardware

• Capable of taking load of data volume

• Have the support of software for required database

• Capable of supporting the communication network

• Satisfy the design specification of system architecture – Central data processing or distributed dataprocessing

8.4.5 Database design : In database design the important issues involved are sub-systems in the organizationand the logic of integration. Technical knowledge of database and knowledge of application systems, theircontrol requirements and designing of reports are essential for efficient designing of database.

8.4.6 Support of top management : To ensure the smooth functioning of MIS top management support isrequired. Top management will support only when they are convinced about the benefit of MIS of the

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organization and confident of efficient performance of processing and regular reporting. Thus, for supportof top management, efficiency of MIS has to be established.

8.4.7 Manpower: Arrangement of right kind of manpower with proper skill is the most consideration forsuccessful operation of the system. Proper planning for training of manpower involved in transactionprocessing and report generation under an MIS system is required to take care of future development ofthe system.

8.4.8 Integration of information: At the time of designing the data bases, provision for integration ofinformation from different sub-systems is essential so that comprehensive information flow can be of greatuse for strategic planning.

8.4.9 Evaluation, maintenance and Control: The effectiveness of an MIS system is evaluated by the capacityof its fulfillment of requirement of information by the management. Evaluation is done by ascertainmentof the views of the users. Maintenance is needed to take care of the gaps, if any, for further growth and forregular smooth functioning of the system. Control means establishment of checks for input data, processingand output to ensure correctness of reports. Proper maintenance and control on effective operation of MISrequired to ensure protection from hazards and smooth functioning on a routine basis.

8.5 Different Approaches for developing Management Information System

For developing, three different approaches are generally practiced which are given below :

8.5.1 Top Down Approach : Top Down Approach starts from the identification of information requirementsof different activities of the organization by the top management in order to have information support instrategic and tactical decision making and designing the information system accordingly. Top Managementprovides the guidelines for basic objectives, policies and plan for developing these sub-systems. In otherwords, this approach designs a model of information flow and same model is used in developing all thesub-systems under the MIS.

Each Sub-systems will have different modules and they are collectively integrated to form a sub-system.The approach of integration is same for all sub-systems. The implementation of different sub-systems isdone on the basis of broad guidelines of the top management. Integration of all sub-systems is done at theend to form a comprehensive MIS for the organization. The implementation process is very scientific,systematic and simple.

8.5.2 Bottom Up approach : In the Bottom Up Approach, each sub-systems for different functional areaslike payroll, Sales Management, Production Management , Inventory Control System are developedaccording to the specifications for each sub-systems on the basis types of input documents, flow ofinformation and output requirements. There is no common approach for system development. Rather, thesub-systems are developed purely on the basis control information requirements for each sub-systems andguidelines generated by the manager of the respective functional areas.

The next step in the bottom up approach is to integrate the information of these sub-systems for acomprehensive MIS for the organization. This step is a complicated one in this approach. The data basestructure of different systems, flow of information and links among them are to be understood thoroughlyto have proper integration. Sometimes intermediate databases are created to collect all relevant informationfrom different sub-systems for integration.

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8.5.3 Integrative Approach : This approach in a more scientific approach for easy integration of sub-systemsand takes care of the limitations in the other two approaches described above by way of better planning.

Under this system, the top management identifies the information requirements from different sub-systemsand specifies other guidelines for integration of these information for effective support to decision making.The managers from different functional areas present the flow of information under individual sub-systems.The aspect of integration of information of different sub-systems is considered at the planning stage. Anymodifications required at different points are pre-conceived at the beginning so that they are taken carefrom the design stage. This approach of implementation allows designing better structure of databasesand ensures smooth flow of information at different levels of management of different functional areas.

8.6 Constraints in operating MIS

Following are the constraints in operation of an MIS :

• Non-availability of experts who can diagnose the requirement of organization and give desireddirection

• Technical knowledge gap between management expert and computer expert

• Non-availability of right kind of cooperation of the employees

• High turnover of MIS expert

• Shortcoming in design leading to non-fulfillment of users’ requirement

• Frequent changes in the information requirement by the users

• Non-standard requirements of management to handle special features of an organization

• Difficulty in quantifying the benefits of an MIS

• Heavy investment requirement for establishment of an MIS etc.

8.7 Limitations

Effectiveness of an MIS depends on its design, software, operating staff and users’ consciousness level.The commonly observed gap between the system expert and the management experts sometimes becomesa bottleneck in designing an effective MIS. Apart from the above, the other limiting factors for MIS are :

• Effectiveness of MIS depends on efficiency of the management in using it and it is not a substitute ofManagement

• MIS only provides information and non-programmed decision making is in the hand of management

• Frequent changes in the information need of management reduces efficiency of MIS

• Success depends on quality of output and their effective use

• Effectiveness is greatly affected where culture of hoarding and not sharing information is there

• Problem of perception of utility

• Lack of proper training and awareness of both operation staff and the users on the use of MIS

• With the change in technology and business environment, management need of information changesthereby suggesting periodic review and continuous monitoring

• Cost Benefit assessment depends on many subjective assessments

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8.8 Information Requirements of Management

Management information system is developed to provide the right kind of information for decision making.Information System, to be truly efficient, has to have interlink with different functional management. In nobusiness, a single function operates in seclusion of other functional areas. Rather, they have close links.Unless different functions are seen in totality, strategic decision making for the organization as a wholecan not be efficient.

Information Requirement depends on the following factors :

i) Operational functions

ii) Type of decision making

iii) Level of Management

8.8.1 Operational functions : The information generated in different functions is different and controlrequirements vary widely. For example, the requirement of information for decision making in productionfunction will not be same as in case of financial function. The type of reports in different functional areasare given in para 8.10.

8.8.2 Type of decision making : Decision making type may be i) programmed decisions and ii) Non-programmed decisions.

Programmed decision making refers to those decision making process which are based on some standardset of procedure established by the management and according to scientific principle of management. Incase of programmed decision making, supporting information sets and reports are standard, well definedand well structured. Naturally decision making process is simple and based on some guidelines. Forexample, stores ledger summary and material consumption reports may help in decision making onInventory control.

Non-programmed decision making refers to those decision making process which does not go by any pre-determined set of guidelines. Normally this type of decision making takes place to handle special businesssituations with the help of experience, judgement and vision of the decision maker. In case of non-programmed decision making, information are unstructured and external environmental information is amust along with internal information sets. For example, for decision on business policy many non-standardinformation like technology change, competitors market share etc is required apart from internal informationof sales of different products.

8.8.3 Level of management.

Information requirement varies with the level of management and purpose. The levels of management inthe order of hierarchy are Top Management, Middle Management and Operational Management. Theactivities of different levels of management are given below :

Top Management : Top management is concerned with strategic decisions like diversification, technologyacquisition, new market exploration, strategic business alliance, takeover, merger etc.

Middle Level : Middle level management is generally involved tactical decision making with the help ofperformance analysis, budget variance analysis, devising better productivity mechanism and control etc.

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Operational Management : Operational Management staff are mainly involved in scheduling the activities,keeping track of progress of day-to-day operations and decisions of well structured problems etc.

The characteristics of information naturally varies depending on the functions of different levels ofmanagement. They are as given below:

Contents of Reports based on level of management:

Characteristics Top Management Middle Management Operational ManagementInformation

Focus of Planning & Strategic Planning Resource Day - to - day activitiesManagement Management

Boundary Internal & External Internal Internal

Volume Summary with Report of Variances Reports of detailedanalysis activities

Integration among Highly integrated Moderately Restricted tofunctional areas integrated functional area

Comparison with 1 to 5 years Previous year or Day - to - daytime frame month to month

Support to decision Relatively Moderately Structured problemsmaking Unstructured structured problems

problems

Level Information type Report contents

Top Strategic information which are relatively • Summary Resultsunstructured and complex. • Comparative figuresExample : • Possible Analytical• Strategic decision making on production presentation

planning, new product, marketing, sales • Guideline for alternativepromotion etc. options

• Planning and Control of different activitiesof the organization as a whole

• Financial Decision making – fundmanagement or resource mobilization

• Business policy decision

Middle Control information which are moderately • Actual performancestructured and less complex . summary and varianceExample : analysis• Tactical Planning • Reports on exceptions• Control information for resource use

and results like weekly sales of differentproducts

• Reasons of variances, if any

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Level Information type Report contents

Opera-ting Supervisory information which are highly • Detailed reportsstructured and simple. • Operational resultsExample : • Maintenance Report• Control on day to day activities like

production, sales, purchases, idle time etc• Scheduling the activities

8.9 Information need for Strategic Decision making

The primary objective of MIS to derive comprehensive information for effective decision making. Theinnovations of Information Technology is to reduce time for acquiring, processing and analyzing data toprovide real time information. Today, business intelligence means accumulation of relevant market relatedreal time information. The success of an MIS depends on the evolving right kind of flow of information forbusiness strategic planning in an organization.

Strategic Information System is to provide best information flow to feed the management in the decisionmaking process to sustain organization’s competitive advantage. Information System with the power ofinformation technology which ensures high speed communication can provide the best support to builddecision making models. The Information System will help the organization in the following ways :

• Re-engineering the business process with fast review the results

• Building Competitive Intelligence through monitoring competitors’ performance

• Capturing Market Information on new technology and products

• Market Trend analysis

• Designing products with the help of CAD tools

Strategic Information System is to provide best information flow to feed the management in the decisionmaking process to sustain organization’s competitive advantage. Information System with the power ofinformation technology which ensures high speed communication can provide the best support to builddecision making models. The Information System will help the organization in the following ways :

• Re-engineering the business process with fast review the results

• Building Competitive Intelligence through monitoring competitors’ performance

• Capturing Market Information on new technology and products

• Market Trend analysis

• Designing products with the help of CAD tools

Types of information are given below.

Critical Success Factors :

• Cost Structure

• Product quality and innovations

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• Customers’ profile

• Customers Services

• Customers’ satisfaction level

• Management development programmes

• Change management and flexibility etc.

External Information set

• Demographic

• Economic

• Technological

• Market measures

• Profit measures

• Industry benchmarks

• Competitors performance etc.

8.10 Different types of Reports

Depending on the level of management, different functional areas will require the following information:

8.10.1. Personnel Management

Top-level (Strategic information)

• Skill information (No of persons in a specialized area with Experience)

• Long term human resources requirement

• Policies on human resources development etc.

Middle Level (tactical information)

• Deployment pattern in different departments

• Personnel Deployment policy

• Performance Appraisal etc.

Operational Level (Operational information)

• Performance

• Leave / absentism

• Punctuality etc.

8.10.2 Production Management


• Policy on production – priority of different products etc.

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• Yearly and monthly comparative results on production of different items

• Information on new technology

• Information on capacity utilization


• Actual performance with target

• Variance and their causes

• Breakdown and maintenance information

Operational Level (Operational information)

• Performance details

• Preventive maintenance schedule

• Machine performance etc.

8.10.3. Sales Management


• Information on new product or new market

• Information on market share

• Analysis of competitors strategy

• Sales growth or fall


• Actual sales – product wise with targets

• Sales Variance and their causes

• Performance of different sales offices

Operational Level (operational information)

• Sales details – branch wise product wise

• Individuals sales personnel performance

• Sales expenses details

8.11 Management Decision making

Today, the managers need information support for decision making. The business activities have becomemore complex and competition has become more acute. For survival, the decision making process must beaccurate and at right time. For this, information base for decision making plays a great role. Developmentin Information Technology have created a new era in management decision making process. The decisionmaking has become increasingly difficult due to following change in the business environment :

• Increase in the number of alternatives to be evaluated

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• Increase in risk in business

• Complex decision making environment

• Time pressure

• High cost of making wrong decisions

8.12 Components of Business Information System

Business Information System comprises of :

i) Transaction Processing System

ii) Management Information System

iii) Expert System

iv) Decision Support System

v) Executive Information System

8.12.1 Transaction Processing System

Transaction processing refers to the processing of information relating to monetary transactions in thebusiness activities like purchase, sale, payment, receipts etc. It is a computer based processing for differentfunctional areas to generate all required reports for day-to-day use in the organization. The transactionprocessing system may be disintegrated or integrated. In case disintegrated transaction processing, dataare collected from respective functional areas and processed and reports are generated for their use. Incase of integrated transaction processing system, an application system which has capability of integratingall functional areas (say, ERP system), transaction processing are interlinked with all data from differentsystem and the reports reflect the impact of integrated information.

Transaction processing may also be in batch mode or may follow on-line or distributed data processingsystem. Example of transaction processing in an organization :

• Payroll

• Accounts Receivable

• Bank Reconciliation

• Purchase Order Processing

• Sales Order Processing

• Inventory Control

• Job Costing etc.

8.12.2 On-line Transaction Processing (OLTP)

On-line transaction processing is carried in a client/server system. In todays competitive environment,information at right time plays a great role in controlling costs of various resources and providing bestpossible services to the customers. In other words, business environment has been characterized by growingcompetition, shrinking cycle time and accelerating pace of technological innovations and companies have

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to focus on better information management. Better information means right information at right time.OLTP are being adopted in wider scale to have the following advantages :

• It can serve multiple users at a point of time

• Technology serves the facilities to collect information from multi-locations

• High flexibility in information processing etc.

In case of Sales and Distribution System in an organization where transactions take place at differentlocations on-line transaction processing is followed to carry out the following basic functions :

• Inquiry handling

• Quotation preparation

• Receiving Order from Customers

• Checking availability of materials / products

• Scheduling delivery

• Monitoring sales transactions

• Invoicing

• Managing Bills Receivables etc.

8.12.3 On-line Analytical Processing (OLAP)

An OLAP software does the analysis of information from data warehouse. The OLAP applications arewidely scattered in divergent application area like Finance Management, Sales Analysis. The real test of anOLAP system is inefficient use of data from databases and computational capability of data to developmodel establishing the relationship of various parameters. In fact, it provides the services of ‘just-in-time’information.

Though OLAP software are found in widely divergent functional areas, they have three common keyfeatures which are :

• Multidimensional views of data

• High analytical ability

• ‘Just-in-time’ information delivery

Rarely a business model limited a fewer than three dimensions. The common dimensions in businessenvironment are organization, line item, time, product, channel, place etc. OLAP system should have theability to respond the queries from a manager within a specified time. The OLAP software must provide arich tool kit of powerful capability of analytical ability.

8.13 Need for Integration of Information Enterprise-wide

The conventional information systems aim at providing information related to different business functionalareas to the respective managers for decision making. The evolution in information system suggest forintegration of information of various functional areas enterprise wide for a comprehensive information setfor more effective decision making. This is possible only by way of taking care of all transactions related

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data from different functional areas on-line and integrating them. Today an Enterprise Resource Planning( ERP) system in an organization grasps all business details in real time and integrate them to see activitiesof various functional areas together so that resources relocation is possible more meaningfully, rationallyand cost-effectively.

Today, many organizations world-wide have clearly perceived the advantages of information integration.A good number of large and medium size organizations have implemented ERP system to derive thegreater benefits of linking all the functional systems for effective use of information.

The following are the needs of integration of information :

• Information for various inter-related parameters provides clear picture

• Comprehensive review of business situation is possible

• Disjoint information may have serious gaps

• Redundancy of information is avoided by scientific linking

• Cross functional impacts in the business is assessed

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8.14 Models used for representing the Information

Iconic scale model : It is physical replica of the system based on different scale from original. Iconic modelsmay appear to scale in three dimensions - such as model of a production process, building, car or anaircraft.

Analytical model : It may be a model for a physical system but the model differs from actual system.Example – Map showing water, mountain etc. by different colours.

Mathematical Model : It represents a data set in the form of graph, picture or frictional diagram. It useshighly mathematical or statistical algorithm to interpret data of huge volume with ease. The algorithmvaries depending on the complexity of analysis of data sets and the type of analysis.

8.15 Analysis of Information

1. Environment Analysis

• Customers – need, choice, disposable income level

• Suppliers – Range of product, pricing, after-sales service

• Competition – pricing policy, quality etc.

• Trade association – industry trend, technology

• Government – control, taxes & duties

• Market – population, demographics etc.

2. Product position analysis

• Market share and growth

• Product performance

• Range of products and level of competition

• Market trend

• Technology and possibility of change

3. Cost Benefit Analysis

• Payback period

• Net Present Value

• Internal rate of return

4. Business Planning

• Operating performance analysis

• Profitability analysis

• Market Research

• Product mix selection

• Budgeting

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8.16 Information requirement for Product Pricing Strategy

• History of change in price

• Demand for the product

• Graph showing demand and price & quality purchased

• Profitability of products

• Price of substitute products

• Price of similar products etc.

Sources of information

• In-house market database

• Industry statistics

• Market research information

• Feedback from customers

• Trade journals

8.17 Executive Information System

An Executive Information System (EIS) is special type MIS meant for top management of an organization.In other words, it is a Decision Support System (DSS) for Executives. Executive decisions are of three types– strategic planning, tactical planning and ‘fire-fighting’.

According to CIMA –

An Executive Information System (EIS) is a set of procedure designed to allow senior managers to gatherand evaluate information relating to the organization and its environment.

Naturally, the EIS takes care the requirement of information depending upon the type of decisions taken atdifferent levels of managers in an organization. In fact, EIS acts as a tool specially designed for differentexecutives to feed their information need in useful formats. A manager can navigate a particular formatwith some amount of computer skill. The EIS is not only limited to internal data source rather facilities toeasy access to common sources of external data is also arranged.

Following are the special features of an EIS :

• It a specially designed tool to feed executives information need.

• It is an easy - to - use and screen based software .

• It provides the executives to facilities of on-line analysis tools like time series analysis, regressionanalysis etc.

• It is not limited to internal data only. Access to external sources of data is also provided.

• It provides the facilities to connect to internet

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• Information is presented in summary format

• It is a comprehensive information system and work in conjunction with DSS.

For an Executive Information System, information requirement varies widely according to the requirementto understand the impact of different variables on the issue. For example, for decision of pricing of aproduct, information requirement may be summed up as follows :

• Recent history of price changes

• Demand for the product

• Graph showing the relationship between demand and price exhibited by recent results

• Effect of demand of changing price over time

• Prices of substitute products

• Price of similar products

• Cost of sales etc.

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Management

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Enterprise Resource Management

9.1 Introduction

Enterprise Resource Planning has become a powerful tool in the h and s of management for effective use ofresources and to improve efficiency of an enterprise. During 1970s Material Requirement Planning (MRP)was a fundamental concept of production management in manufacturing industries. In today’s rapidlychanging business environment every organization has to face new market, new st and ard for qualityassurance, new competition, increasing customer expectations. As a result the business enterprises are in aconstant need of reviewing and re-engineering their processes in order to survive and grow undercompetitive environment. With the advent of innovations in information technology, a concept of integratedapproach embracing all functional areas have been evolved. This has led to development of ERP packageswhich were originally targeted at manufacturing industries and consisted mainly of functions like SalesManagement, Production Management, Accounting and Financial Affairs. In the recent years this hasbeen extended to all industries covering whole management functions like:

• Manufacturing

• Material Management

• Quality Management

• Sales and Distribution

• Logistic Management

• Maintenance Management

• Human Resources

• Finance

• Strategic and Operational Planning etc.

Some popularly known ERP packages:

• SAP AG

• Oracle Corporation

• Peoplesoft

• BaaN

• J D Adwards

• INTENTA (Sweden)

• QUAD

• Marshall (Developed by Ramco System, Chennai, India)

• SSA (System Software Associates, Inc., USA) etc.

ERP Package:

It is a software with the help of Database Management System integrating information related to allfunctional areas. Globally acceptance of ERP System is in great dem and . Industry analysts are forecasting

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growth rate of more than 30% in next five years. The reason for accepting ERP System replacing their oldbusiness system are as follows:

• Improved business performance through optimum resource utilization

• Reduction in manufacturing cycletime by integrated planning process

• Better support Customers in fast changing in market conditions

• Better Cost Control mechanism by way of accurate costing system

• Enhanced efficiency in control through feedback information and online access to accurate information

• Establishment of Decision Support System etc.

Three fundamental characteristics of an information system are accuracy, relevancy and timeliness. To-day, time is a crucial parameter for good quality service to the market. In fact, there are interlinks amongall the systems. A data in isolation does not give right direction in decision making. When the impact of thesame can be seen in all activities in totality, the information becomes far more relevant. A well designedERP package can provide these advantages.

9.2 Selection process of an ERP Package

There are many ERP packages available in the market. Analysing all the packages for choosing the rightone is a time consuming process. Thus, it is better to limit the number of packages at the beginning for thepurpose of evaluation. Looking at the product literature of the vendors, one can eliminate the packagesthat are not at all suitable. Normally this evaluation process is done by a Committee. What is required to bedone is gap-analysis between the requirement of the company and capability of the package. Presentationor demo from the selective vendors will provide some direction towards choosing the best. Of course, costof the package is also a key factor. Cost benefit analysis is also to be done.

The Common Criteria for selection for a package:

1. How best the package fits the requirement of the company

2. Provision for accommodating the changes in the system

3. Implementation and Post Implementation support from vendor

4. Reliability of Vendor

5. Change in Hardware and Skill requirement

6. Cost of the Package and Budget

Cost of ERP Implementation

A budget is required for implementation of an ERP package. It is not only the cost of ERP package but alsothere are many hidden costs that are to be considered. The following costs are to be considered:

1. ERP package cost.

2. Consultant cost.

3. Cost of Data conversion.

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4. Cost of training.

5. Cost of testing.

6. Cost of Post-Implementation support.

Business Process Re-engineering and Implementation of ERP

According to Peter F. Drucker, “Re-engineering is new and it has to be done.” Business Process Re-Engineering (BPR) aims at performance improvement through dramatic change in organizational structure,skill development, change in technology and change in mindset of people. BPR is a process which exploresthe possibility of doing things in different ways to improve efficiency in the operation and it involves re-thinking, renovation, redesigning, retooling. In other words, it is an exercise towards transformation of anorganization to dynamic change which involves lot of innovations in every activity.

Implementation of ERP involves replacement of all existing Information Systems. Thus at the beginningthere is a need to change the traditional method of data gathering and analysis because data from differentdepartments needs to be integrated. The question of performing in isolation vanishes. Developing linkageof data from different sources, correlating them and understanding the impact of an integrated system arethe important processes involved in implementation of an ERP System. To device a system with holisticview of enterprise requires integration of isolated piecemeal Information Systems and facilitates seamlessflow of information across departmental barriers. In other words, implementation of ERP Systems involvesgreat change in information flow and reporting system. The success depends on support and co-operationof all users and managing the changes.

BPR exercise is essential for getting true benefit out of an implementation of ERP package. The main reasonis to bring the changes in the physical system, data flow and mindset of the people to effectively utilize itscapability of integrating information for the benefit of management decision making. BPR exercise may bedivided into different phases as given below:

Phase I: Begin Organizational Change

• Understanding the current state of Organization

• Assess the need for changes

• To explain how to bring the changes

• To project the new dimension in the business

• To make a document on new organizational structure

Phase II: Bringing Re-engineering process

• Establish BPR Organizational structure

• Selection of people to be involved in BPR

• To fix the responsibilities of these people

Phase III: Identify BPR Operations

• To pinpoint the processes to be re-engineered

• To assess the potential changes

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• To fix priorities of activities

• To draw performance matrix

• To establish business strategies

• To identify the potential barriers

Phase IV: Understanding the existing process

• To Develop the model of current process

• To understanding the technology currently in use

• To assess the present flow of information

• To identify the gaps in the flow of information

Phase V: Newly designed model

• To draw new work flow

• To define new process steps

• To describe the new information requirement

• To assess new technology requirement

• To assess skill requirement

Phase VI: Transformation

• To develop migration strategy

• To devise action plan

• To reallocate workforce

• To develop training curriculum

• To ensure smooth change over to new process

Business Modules in an ERP Package

All ERP packages contain a set of modules. These modules are related to different functional areas likeFinance, Manufacturing and Production Planning, Materials Management, Selling and Distribution andso on. These modules in a business systems have close relationship. An ERP system takes care the flow ofinformation from different modules and understanding the interactions among them. To avoid dataredundancy, data generated from the activity centres are entered and the same data is considered bydifferent modules of the system where the same is relevant for the purpose of understanding the impact ofone the other. This is the basis of integration of information. This is the most important feature of an ERPsystem.

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Module of ERP Package

1. Finance Module: Finance module in an ERP system will have the following sub-systems :

• Financial Accounting: General Ledger, Accounts Receivable, Accounts Payable, Fixed AssetsAccounting etc.

• Investment Management: Investment Planning, Budgeting, Depreciation, Forecast, Simulationetc.

• Controlling: Overheads Cost Controlling, Activity Based Costing, Product Costing, ProfitabilityAnalysis etc.

• Treasury: Cash Management, Treasury Management, Market Risk Management, FundsManagement etc.

2. Manufacturing Module : Manufacturing Module generally has the following sub-system:

• Material and Capacity Planning

• JIT/ Repetitive Manufacturing

• Engineering Data Management

• Cost Management

• Quality Management

• Configuration Management

• Tooling etc.

FinancialManagement

HumanResource

Sales andDistribution

Manufacturing

Management

MaterialManagement

QualityManagement Maintenance

Management

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3. Human Resources Module: Human Resource Module generally has the following sub-systems :

• Personnel Management – HR Master data, personnel administration, Recruitment, Deployment,Transfer etc.

• Organizational Management – Job Specification, Staffing Scheduling, Personnel cost planning etc.

• Payroll Accounting – Salary Calculation, Income Tax Calculation, Accounting for Fringe Benefits

• Time Management – Staff Planning, Work Scheduling, Time Recording, Absence recording etc.

4. Material Management Module: Material Management Module generally consists of the followingsub-systems:

• Material Procurement planning

• Purchasing

• Vendor Evaluation

• Inventory Management

• Material Inspection etc.

5. Sales and Distribution Module:

• Master Data Management

• Order Management

• Warehouse Management

• Shipping

• Billing

• Pricing

• Sales Support

• Transportation

• Foreign Trade etc.

ERP Implementation Life cycle

ERP implementation project involves different phases which have definite activities as explained below:

1. Pre-evaluation screening – to search for perfect package which will be most suitable in terms offunctional fit of the business process, skill set available and easiness to adopt.

2. Package evaluation – to understanding the performance of the business and do the cost benefitanalysis.

3. Project Planning Phase – to make a tentative plan for implementation in terms of time, identificationof person responsible for co-ordination of the implementation programmes, skill development andmonitoring the progress.

4. Gap Analysis – to identify the gap between the existing system and future expectations from the ERPsystem so as to optimize the outcome from the implementation programme.

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5. Re-engineering – to bring the necessary changes in the system in terms of physical system, hardware,mindset of the people, data flow and reporting.

6. Configuration – to install the necessary hardware, data base management system and configurationof the ERP system accordingly.

7. Testing – to test the system with test data set like entering data, validating them and generatingreports for all modules and checking their correctness.

8. End user Training – to impart training to people from different functional areas who will be associatedwith operations and handling reports.

9. Going live – to finally switching over to new system with data migration, and running the systemwith live data of all functional areas.

10. Post – Implementation – to arrange for maintenance of the system in terms provision of technicalexpertise in cases of problems.

Implementation Methodology

The methodology of implementation of ERP system involves different activities like understanding theneed for the system, benefits to be derived, people to be associated, need for their skill updates, databaseconfiguration etc. Generally, an ERP implementation can be divided into four phases:

1. Understanding the problem

2. Defining solutions

3. Undertaking technical work

4. Going live

5. Post implementation maintenance

1. Understanding the problem: This is the phase where in-depth analysis of business system is undertakento understanding the need for an integrated system in handling data and how to improve the dataflow towards better management reporting and effective decision making. The company goes forevaluation of different packages and selects the one which is the perfect fit to the requirement. In theevaluation process of the package the following parameters are taken into consideration:

• Functional fit with company’s business process

• Degree of implementation among various modules

• Flexibility

• Technology requirement

• Cost of the system (Systems, Training and Maintenance)

• Time of implementation

• Skill requirement

• Technical support

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2. Defining the solution: This phase considers all the intricacies of the systems so as to take care of allthose during the development process. The gaps in the present system, scope of improvement in thedata flow and need for integration for better control are the most important three areas which arecritically evaluated to arrive at a solution set. Finally, the outcome of the same is taken into considerationfor software. This phase gives solution for data migration process, volume of data and database size,plan for implementation in terms of technical requirements and timeframe. In fact, in this phase,project team is built, project schedule is defined for all broad activities.

3. Undertaking Technical Work: In this phase, the activities are broken into details. The work involvedin this phase are:

• Procurement of hardware

• Configuration of the system

• Data migration

• Developing and testing the interfaces

• Developing the new procedure associated with the system

• Testing the new system

• Training the end-users

4. Going Live: In this phase, the system is to be finally implemented in new environment with real lifedata set and to the satisfaction of the end-users. In ERP systems, the integration of all the modules isthe critical part. The end-users must understanding the sequence of operations, how one moduleinteract with the others and what are the restrictions in operation in terms of priority so as to establishproper checks at all levels in the process. The co-ordination among project members for differentmodules is very essential for smooth and successful implementation.

Post-implementation maintenance: Once the implementation is over, the services of vendor and thehired consultants will not be available. Trained in-house employees may have limited exposure totake care of all the problems just after implementation. Post implementation needs a different set ofroles and skills to solve the problems in an integrated system. The training will never end. Newfunctionality may be added which will invite different technical problems like enhancement of system,fresh configuration for added integration features. Thus, this is a very critical phase. To reap the fullbenefit of ERP system, there should be arrangement for continuous training of employees and periodicalreview on how to enhance the advantage from the system.

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Information System Control

10.1.1 Risk in Information System:

Risk is the probability of happening something adverse. An information is vulnerable to many risk factorswhich are categorized below.

Risk Management is a process of assessing risk and reducing it to an acceptable level. Steps involved in it are:

• Understanding the sources and causes of risk.

• Collection of data related to risk to analyse their nature and frequency.

• Evaluation of magnitude of risk.

• Develop policy to minimize the risk.

• Develop methodology to prevent their occurrence.

• Implementation of the methodology.

10.1.2 Different Categories of Risk Factors:

1. Hardware

• Damage by lightening, fire, earthquake etc

• Failure of communication and network system

• Power failure

• Breakdown of computer system etc

2. Software

• Errors in programs

• Errors in operation

• Weakness in control

• Damage by computer virus etc

3. Database

• Destruction of data-file

• Loss in data integrity

• Misuse of database etc

4. Input Data

• Incorrect data

• Damage of data file

• Collusive fraud etc

5. Operational Risk

• Failure of machine

• Failure of software

• Wrong operation etc

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6. Others

• Technological obsolescence

• Fraud

• Damage by disgruntled employees

• Unauthorised access by hackers

• Damage by natural disaster like fire, flood

• Leakage of sensitive information etc

10.1.3 Developing Methodology to Minimize Risk

To create proper environment, to establish proper Control Mechanism and regular vigilance on the properfunctioning of the established norms in the Information system are essential to reduce the risk. StandardAuditing Practices (SAP-6) emphasise the need for proper internal controls to check the probable hazardsfrom outside.

Following provisions and environment are developed to check the occurrence of risk factors or to take carethe eventuality in case of failure:

i) Documentation: Documentation on Policy, procedure and System Administration.

ii) Creation of congenial Environment: Care Environment free from hazards to minimize chance of damageof systems.

iii) Preventive Maintenance: Regular and timely preventive maintenance.

iv) Quality Management: Ensuring the quality of software at the development, testing and operationalefficiency etc.

v) Security Management: Strict control on unauthorized access by password and different types ofpermissions to allow access to / operation of the system to different sets of operating staff.

vi) Recovery System: Regular back-up of data files to take care of damage or failure and parallel processingfor real-time system.

vii) System Administration: Follow of strict code of system administration.

viii) Set of Routine Procedure: Keeping regular vigilance on probable damage options.

ix) Training: Training to improve the awareness of control among the employees.

x) Ethical Standard: Special care for maintenance ethical standard in the working environment etc.

10.2 Information System and Management Control

In an organisation it becomes the management responsibility to develop its information system to attainits objectives. Accordingly, it has to structure its organization of Information Systems. The basic objectiveof information system is to provide the required smooth flow of information for planning and controlactivities. To strengthening the internal control system, Standard Auditing Practices (SAP-1) suggest thefollowing steps:

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i) Framing policies and procedure for Effective Control.

ii) Establishment of sound organization structure with clearly defining responsibilities and authorityfor access to information, supervision and control on information System.

iii) Creation of better control environment through effective budgetary control system

iv) Ensuring smooth flow of information for effective control.

v) Regular Monitoring the internal control systems.

vi) Periodic review of control system.

10.2.1 Policies and Procedure for Effective Control:

Policy establishes the rule of procedure and boundaries for authority of individuals in action and decisionmaking. Policy for Internal Control for Information System basically aims at fixing sequence of flow ofdata, taking measures for security of information, ensuring timely processing of data and generation ofInformation Reports and use of the same with definite objectives. Procedure will delineate the detailedsteps to be followed for execution and check the deviations, if any. Policy will depict the guidelines on thefollowing :

• Level of authority

• System of cross check

• Procedure of monitoring

• System of review of control system etc.

10.2.2 Creation of Proper Processing Environment

Creation for proper environment is essential for effective control system. The two important componentsthere are of great importance-discipline and ethics. Unless these two are enforced no control system can befull proof. Control is done to see that the existing procedure is running smoothly and with comparisonwith some standard. Performance budget must be prepared honestly. If the budget is vague and sub-standard, all control measures are futile. Thus creation of environment with professional ethics must betaken care properly. Without proper environment, no amount of deviation from norms get properly checked.Internal Control system in Information system aims at :

• Data Security

• Preventive maintenance

• Proper resource management

• Timely data processing

• Cost of data processing etc.

10.2.3. Smooth Flow of Information:

Success of an information system depends on its timeliness and quality of report. The user should appreciateits utility of the information report. Utility of report is great when user is fully dependent on it. To ensurethese two intergradient in flow of information, the following checks must be built in into the system:

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• Documentation

• Well defined responsibility

• Clearly spelt-out duties of individuals in Information Systems department

• Preventive measures against failures

• Standardisation in the system

• Proper user manual

• Good training

• Regular check procedure etc.

10.2.4 Monitoring:

Regular supervision is a preventive measures for smooth functioning of the system established with anobjective. This is extremely important in data processing. The fraud uses the loose ends in the system. In antightly controlled environment, the chance of fraudulent activity or sabotage is greatly reduced. In dataprocessing fraudulent activity or sabotage may cause great damage. Thus, regular supervision of machine,machine usage, input and output with proper records with signature of proper authority should be an in-build system in Data Processing Environment. Statistical records of deviations in flow may help to reviewthe system.

10.2.5 Periodic Review

Periodic review of the control system is essential to assess the efficiency of the existing control system,their adequacy and change needed to match the change the environment and technological developments.Review report should be submitted to senior management and independent observer to have independentassessment of the justification of the suggestions under the review.

10.3 Controls in Information System

Now, application of computer in a business organization has become wide ranging. Sometimes it happensthat the computer use in different application systems have started in different points of time and that toonot in a planned manner. The problem becomes more when the approach is half-hearted and withoutsatisfying the technical requirements.

With the hardware cost going down day by day, the online processing is gaining tremendous popularityin the data processing environment. Today, data base management system has brought the idea of dataintegration to reduce data redundancy and effective combination of information in reports to derive betterdecision making process. This development has brought the need of data control more seriously thanbefore. A familiar expression is there – GOGO (garbage-in-garbage-out). Thus, control of both input andout put has become more than necessary to ensure correct reports, avoid chaos, to enhance effectiveness ofcomputer based application system.

Controls in Information System mean the policy, procedure and system followed to ensure the desiredobjectives.

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10.3.1 Organization Structure Control:

Each operational task must be well defined. Segregation of tasks and responsibility of execution of theeach task should be assigned to definite group of personnel to fix accountability. Following tasks are to aassigned to a designated group of personnel as given below:

Task Staff Responsible

System Development Systems Analyst, Programmer

Computer Operation Computer Operator

Database Administration Database Administrator

Hardware Maintenance Maintenance Engineer

Network Maintenance Network Specialist

Data Security Input/Output Control Staff

10.3.2 Control in Operation Environment

The control and monitoring on the following things essential to provide suitable environment for dataprocessing :

• Proper earthing of electrical line.

• Provision of UPS to protect the system from abrupt failure.

• Maintenance of proper register for users.

• Cleaning by vacuum cleaning.

• Regular electrical maintenance.

• Diagram of Electrical/network cabling to be documented.

• Allowing use of only licenced version of software.

• Provision of Annual Maintenance of Systems etc.

10.3.3 Preventive Measures against Damage:

The Preventive control measures are to protect the system from accidental damage or intentional threat.These may be summarized as below:

• Proper security measures against theft and damage of hardware.

• Arrangement from disaster recovery.

• Proper back up system to prevent from accidental data loss.

• Routine check on unauthorized access to system.

• Prevention from damage of Operating System, Programs and data files by virus.

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10.4 Controls in Processing Environment

Some controls during data processing is required to ensure that the data which will be processed arecorrect. For this purpose, input control, access control, process controls and output controls are necessary.Details of these controls are described. A summary of these control are given in the table below:

Category Control Purpose

Input Control Input Coding To ensure correct codification of data item. This is done validitychecking of codes in the program.

Record Count To number every records to detect any missing records

Hashing Use of codes in place of name of person or party to avoiddisclosure of confidential information.

Batch Total Putting manual total of a batch of records to match the samewith computer batch total to avoid mistake in data entry.

Authentication To ensure authenticity of the input.

Access Control This is done by the use of password or other restriction toallow access to data, file or devices.

Processing Format Check To check the validity of data size and data type in terms ofControl alphabetic and numeric.

Limit Check To ensure correctness of figure by a check of logical limit ofvalue of single transaction. For example, limit for a cashtransaction may be Rs 5000.

Control Total Control total is generated by the computer during processingto allow checking with the input control total.

Exception To identify the input data which are logically invalid.Report

Output Control To check the output control figures generated by machine withthe corresponding input control figures.

Audit Trail To follow the processing sequence of an input data with thehelp of Reference No. corresponding to a transaction. This isa checking process by the auditor with the help of test datawhether a transaction is processed in the right sequence toreflect its impact correctly in the reports.

10.4.1 Input Control:

Input control is required to ensure correctness of data from source and their completeness and authorization.Errors in input data may be out of following reasons:

• Incorrect data out of clerical mistake.

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• Wrong codification.

• Willful mistake for doing fraud.

• Mistake at the time of data entry etc.

Following are the checks must be followed to ensure accuracy in input:

• Pre-numbered Source Documents - Printed source documents with numbers are only documentsallowed to prepare source documents to have adequate physical security on the document preparationand to prevent fraud.

• Input Authentication Check – to see whether input documents are authenticated by well definedofficial of user department.

• Batch Control – to validate the correctness of important mathematical information of a group ofdocuments with its manual batch total and computer total.

• Check Digit – check digit is a redundant digit derived from some mathematical relation out of otherdigits of the code which is incorporated in the code itself to ensure correctness of code.

• Proper system of verification of data entered in the computer–off-line/on-line.

• Check List – detailed list of data entered with indication of logical errors or errors in codes.

• Parity Check – a system of adding an extra bit to each character in order to check the possibility ofloss of bits during data transfer/transmission to media like tape.

10.4.2 Access Control:

Access to computer should be restricted to authorized persons only to prevent fraud, damage and ensurebetter security of information and assets. For this purpose, a security policy and framework must be devised.The primary control for security purpose is access control. Access control is executed with the followingchecks :

• Permission to level of access to system depending on category of user.

• Periodic change of password should be made mandatory.

• Password of System Administrator should be kept in safe custody under sealed cover.

• Access restricted to only users presently working in the Department/Branch.

• Use of Standard Anti-virus software in all PC’s.

10.4.3 Processing Control:

Processing controls means checking the logical errors of input data and ensuring validity of data accordingto the rules of source document preparation. They are :

• To check the validity of data size and data type in terms of alphabetic and numeric data

• To ensure correctness of figure by a check of logical limit of value of single transaction. Forexample, limit for a cash transaction may be Rs 5000.

• Control total is generated by the computer during processing to allow checking with the inputcontrol total.

• To identify the input data which are logically invalid

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10.4.4 Output Controls:

Output Controls are to ensure that the reports generated for the users are correct and the output filescreated or/and database updated are correct and consistent. Other controls are:

• Checking the output reports with batch control totals.

• Thorough checking of correctness of output reports.

• Distribution of output to the authorized users only.

• Back up of output files at regular intervals.

10.5 Documentation Control

• All manuals — System Manual, User Manu both for hardware and software must be kept in thedepartment.

• Logbook must be maintained for processing records and generation of reports with signature ofauthorized person.

• Records of usage, down time, maintenance calls and maintenance service must be maintained.

10.6 Computer Fraud

Computer fraud is an illegal action with the help of computer technology to make financial gain, to haveunauthorized access to private information of others, to damage software/data etc.

The computer frauds have become a very sensible issue because of the following situation:

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• Wide interconnectivity among the machines through network thereby making information systemmore vulnerable.

• Internet and e-commerce have increased the risk of hacking.

• Growth of e-cash transactions will induce more frequent attempt for interception.

• Common tendency for piracy of software.

• Strategic damage out of unauthorized access by competitor/enemy to confidential information.

Some common computer frauds:

Technique Description

Hacking Unauthorized access to software and information.

Cracker Unauthorized access to machine with intention of damage to data,software etc.

Password cracking To have an access to systems resources by decrypting password.

Software piracy Copying software for illegal use without paying due price.

Virus Use of software to damage set of software/data files in the machine.

Trap door Access to system by passing normal systems control.

Super-zapping Access to special system programs bypassing normal systems control.

Measures for detection of fraud:

• Proper audit at regular intervals on control measures.

• Use of expert software to provide advance alert signal etc.

Measures for prevention of damage out of fraud:

• Proper security measures.

• Proper audit at regular intervals on control measures.

• Disk imaging.

• Keeping regular backup of important files/databases.

• Train employees in measures against fraud prevention.

• Identify and manage disgruntled employees.

• Punishment to unethical activities.

• Use of only licenced version of software etc.

10.7 Security of Information System

Security of information system refers to protection of databases, output reports and assets from damage.Damage may be from theft, sabotage, natural disaster, fraud etc. The security of information system meansprotecting the confidentiality, availability and integrity of information. According to IFAC guidelines –

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information security is for protecting the interest of those relying on information, and the systems andcommunications that deliver the information, from harm resulting from failures of availability,confidentiality and integrity.

In today’s scene of global competition, the information system for an business organization is the pivotalpoint on which business run. Security has become of a big concern for wide scale of network use,development of innovative techniques of frauds and damage by unauthorized access. The damage are offollowing types:

• Leakage of sensitive information by unauthorized access.

• Misuse of system.

• Fraud by using loose controls or with the aid of technology maneuvering.

• Theft or physical damage of computer, software etc.

Objectives of Information Security:

• Assignment of responsibility for maintenance of data.

• Standardization of system of access to prevent damage by unauthorized access.

• Safeguard against threat of loss or damage of hardware, data and software.

• Arousing awareness on risk among the staff.

• Protection from chaotic situation out of system failure etc.

Role of Security Administrator:

The threats of damage or loss of hardware, data and software by frauds have evolved a newresponsibility in information system environment, that is to ensure the safety to information system.His responsibilities are:

• Framing information system security policy.

• Devising measures to ensure safeguard from possible threats.

• Train the staff for effective action.

• Implementation of security system.

• Monitor the security measures.

• Developing action plan for emergency recovery.

The security measures have become prime important to minimize the risk in safeguarding the loss ofassets and information of the organization. Following measures are taken to safeguard Computer equipmentand databases:

• Arrangement of physical security

• Access Control

• Insurance against damage

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10.7.1 Physical Security

The following physical restrictions to unauthorized access are termed as physical security:

• Door

• Machine Lock

• Guard

• Floppy disk access lock etc.

10.7.2 Access Security with Checks in Computer System:

• Password

• Biometric security

• Firewall

• Database security

• Network security

10.7.2.1 Password Security:

This security measure is to control access to computer or system with the help of password. The systempreserves the password in encrypted manner. Password is a confidential key to a person and in normalsituation no person is in a position to have the password. It is an easy and simple system of security controlmeasures to prevent access to a computer/software.

10.7.2.2 Biometric Security:

Biometrics is a mechanism of defining user profile based on physical parameters and behaviour. Some ofthe biometric characteristics are signature, voice, facial scan etc. Authentication with the help of biometriccharacteristics needs special hardware and software for image processing. The identifications/authentications are stored in databases which are checked at the time of allowing access.

10.7.2.3 Firewall:

Firewalls offer an effective system to protect access by unauthorized user from outside. The main featureof firewall is packet-filtering router so that vital information does not pass to any unauthorized intruder,even if he manages get access to the network system. It is a system of security in the network with the helpof hardware and software. A software checks all incoming and outgoing internet traffics. The firewallroutes the massages to a safe area to avoid any danger in the in forward transmission of messages. Thescreening by firewall software may delay the transmission process but ensures proper security.

Limitations of Firewall

• Passing on information by internal employees through internet can not be checked.

• Firewall can not protect the system from virus.

10.7.3.4 Database Security:

Access to database is controlled based on the degree of sensitivity of the data in the database. Users arecategorized on the basis of permission to be given for access to different levels of data on the basis ofsensitivity and control is made by the database administrator.

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10.7.3.5 Network Security:

Access to network is controlled by password, firewall security measures. Network protocol also takes careof security aspect for access to network. Apart from these, Terminal Access Controller Access ControlSystem (TACACS) and Remote Authentication Dial In User Service (RADIUS) also have remoteauthentication mechanism to control access.

10.7.3.6 Insurance:

Proper Insurance coverage of assets and software is essential to protect the financial loss to the company .It may cover damage/loss to

• Computer

• Database

• Software

• Media etc

10.8 Disaster Recovery

10.8.1 Disaster:

Disaster means sudden great misfortunate happening which can not be prevented. Disaster may be of twotypes – i) Natural like flood, earthquake, hurricane damage etc. ii) Technological like failure of computer,electrical fire etc.

10.8.2 Disaster Recover Plan:

The damage under disaster is generally enormous. The question of recovery in case of disaster comes fromdata. Data may be categorized as critical, vital, sensitive and non-critical. Recover plan may be devisedaccordingly to given priority of recovery of data of different importance.

Emergency Action : In the first stage the notification of damage is to be given to the appropriate agency/authority like fire service, police, insurance company etc. Then following action may be taken dependingin the situation to save personnel, equipment, data etc like:

• sounding alarm bell

• use of fire extinguisher

• saving the back-up of software, data etc.

Recovery Action: There needs an advanced planning for recovery of data under disaster. Generally, thedisaster recovery planning is done by a Recovery Committee and execution of recovery programmes isdone under its supervision and control. These are:

• Backup Application software and backup of databases at a regular interval to be preserved insome other location.

• Mirror imaging of disk.

• Selection of alternative computer system.

• Restoration of application software and databases in the new computer system.

• Critical evaluation of performance of the application software.

• Assessment of loss of databases.

• Plan for recovery of data loss etc.

InformationSystemAudit

STUDY NOTE - 11

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Information System Audit

11.1 Objective

An organisation should control and audit its computer based information system. Otherwise, the costs oferrors and irregularities that may happen in these systems may prove to be unaffordable. Uncontrolleduse of computers can have an adverse effect on society. For example, inaccurate information may cause amisallocation of resources within the economy and there may be fraudulent practices causing harm to thesociety – mainly to the low income group general public. Because computers play a vital role in processingdata and also in the decision - making process, it is imperative that their uses must be controlled. Thereasons for such measures are:

» there may be adverse consequences as a result of losing the data resources,

» misallocation of resources because of incorrect decision based on incorrect data,

» computer systems being uncontrolled, there is possibility of computer abuse,

» the high value of computer hardware, software, and personnel,

» high costs of computer error,

» the need to maintain the privacy of individuals persons, and

» the need to control the evolutionary use of computers.

11.2 Definition

The information systems auditing is the process of collecting and evaluating evidence to determine whethera computer system safeguards assets, maintains data integrity, allows organisational goals to be achievedeffectively, and uses resources efficiently.

Thus the information systems auditing supports the following objectives:

• safeguarding of assets which include hardware, software, people i.e. knowledge, data files, systemdocumentation etc.

• data integrity i.e. completeness, soundness, purity and veracity,

• system effectiveness i.e. it has knowledge of user needs and facilitates decision making process in theorganisation,

• system efficiency i.e. use of minimum resources to fulfil the desired objectives, and

• statutory compliance i.e. rules, regulations, or conditions to be complied with under various Acts,Laws, Regulations etc.

11.3 Internal Control

The basic objectives of information system auditing are asset safeguarding, data integrity, systemeffectiveness and system efficiency. But all these objectives cannot be achieved unless there is a system ofinternal control imposed by the management. While the conventional approach towards internal controldo not change under a computerized environment, but the implementation of the control system mayhave to be adapted to the functioning of the computer system. Major areas of internal control system arediscussed hereunder:

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• Separation of duties and responsibilities: In a conventional system, responsibility is separated foreach of the aspects of initiating transactions, recording transactions and maintaining custody of assetsin order to prevent or detect errors and irregularities. In a computerized system, this may not bepossible in a straight manner. Duty or authority to change program/data may be separated from thepoint of view of control.

• Delegation of authority and responsibility: This is an essential control both under manual andcomputer system. In a multi-user database management system, the control problem regarding integrityof data and redundancy of data is minimized. Where more users have authority for developing,modifying, operating and maintaining their application systems instead of by computer professionals,there may be some difficulty in exercising control.

• Specialist/trained personnel: If the organisation maintains professionally qualified specialist personnele.g. System Analyst, Database Administrator, Application Programmer, Computer Operator etc., thecontrol aspect becomes easier to be exercised. Many companies do not maintain such specialists becauseof high cost of retention. In that case competent and properly trained personnel should be deployed,as the power vested in the personnel responsible for computer system generally far exceeds the powerexercised by the personnel engaged in a manual system.

• Authorisation system: Authorisation may be of two types viz. general authorisation (e.g. authority tosell goods at a price as per approved price list) and specific authorisation (purchase proposal requiresvetting of the appropriate tender committee depending on type of tender/tender value). Whileevaluating the adequacy of authorisation procedures, auditor should examine both the work of therelated person and the specification of the program processing.

» Audit and management trail: Computer system should be so designed as to provide audit ormanagement trail by maintaining a record of all events and providing access to any record. Adequateaccess control and logging facilities will ensure preservation of an accurate and complete audit trail.

» Physical control of assets and records: In a computer system, because of concentration of informationsystem assets and records, risk of losses due to computer abuse or a disaster increases. Therefore,backup or recovery procedure should be subjected to audit.

» Adequacy of management control: Under a client-server system and communication network in use,operational staff and managerial personnel may be separated by distance and control is to be exercisedremotely. Therefore, parameter of control should be in-built in the computer system to enable adequatecontrol which otherwise would have been exercised through physical supervision, observation andinquiry under a manual system.

» Performance control: The controls incorporated in the program development, modification, operationand maintenance should undergo evaluation to ensure veracity of program code.

» Comparing recorded data with assets: The recorded data and the assets represented by the datashould be compared periodically to ascertain any incompleteness or inaccuracies in the data.

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11.4 Audit Planning

Managing information system audit function is based on the five traditional management functions viz.planning, organising, staffing, leading and controlling. Furthermore, the impact of professionalism andchanging technology should also be taken into account. In view of outsourcing trends in industries, themanagement of an internal audit group as well as an external audit group should be kept in view dependingon circumstances.

11.4.1 Planning Function

In managing the information system audit function, two types of audit plans are envisaged :–

• Long - run plans

• Short - run plans

The objectives for long run planning are to:

• formulate overall direction for the function,

• make provision for adequate resources to discharge the responsibilities effectively and efficiently.

The determination of the former objective will have its consequent effect on the latter objective.

Long run planning of the audit function depends on examining the criticality or relative importance placedby the organisation on the four goals viz. asset safeguarding, data integrity, system effectiveness andsystem efficiency. Although the four goals are existent in all organisations but their relative weightagecannot be same e.g. asset safeguarding and data integrity in a stock exchange cannot be of same importanceas in a consultancy firm. The nature of and extent of audit planning are also guided by the importance ofthe existing information system vis-à-vis the future information system in the organisation. The extent oflong run planning needs increases with the increase of the importance of future information system.Extensive short - run audit planning is necessary if importance of current information system is high.

The objective for short - run planning is to undertake a risk management program to enable systematicevaluation of exposures faced by the organisation and reduction of such exposures to an acceptable level.

Risk management should be proactive rather than reactive.

A risk management program enables to identify the area for more detailed work associated with audit ora security program and overall control of the level of audit risk associated with the organisation. It wouldthen be possible to determine the resources required to conduct the program of audit.

11.5 Documentation

The documentation required to ensure functioning of information system in an organisation are of thefollowing types:

• strategic and operation plans

• complete system and software documentation

• manuals

• memoranda, books and journals

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A part of the documentation may be in magnetic form e.g. where systems are developed with the help ofCASE (computer aided systems engineering), the data-flow diagrams, entity-relationship diagrams etc.may be in machine-readable form, some documentation may be on optical disk (CD-ROM).

Effective management of information system documentation may be difficult for the following reasons:

• responsibility for documentation may be dispersed throughout the organisation (e.g. it may becentralised to support mainframe/minicomputer system but may be vested in all users ofmicrocomputers).

• documentation may exist in various forms at various locations viz. in hard-copy form, in magneticform, in microform or in some combination of forms.

• distribution of documentation to many users.

• ensuring documentation to be up-to-date at all locations and controlling access to authorised usersonly.

• maintaining adequate backup for documentation.

• managing inventory of acquired software and licensed software.

Depending on the importance and span of control of documentation, it may be necessary to maintain aprogram library under a documentation librarian to ensure –

• secure storing of documentation

• access only to authorised personnel

• documentation to be up-to-date

• adequate backup

• managing of inventory of acquired/licensed software to prevent loss, illegal copying, non-complianceof terms and conditions of licensing agreement, unnecessary purchase of additional copies of software,improper maintenance of backup. Help of report generating utility or software may also be taken for review.

Auditors by way of querying, observation and reviewing documentation should evaluate the efficacy ofthe system. Auditors should be satisfied that documentation is maintained securely and it can be accessedonly by authorised persons. The objective is to prevent improper use of or modification to software andensure availability of documentation for maintenance of the software.

11.6 Testing

The purpose of testing is to determine whether the developed or acquired software achieves its specifiedrequirements. During this process, program design errors or program coding errors may arise. Eveninsufficiency or inaccurate specifications may be detected. Testing can however detect only presence oferrors and not absence of errors. Designing of tests is very important for elimination of errors. It involvesseven steps as under:–

1. Selection of the boundaries of the test – whether a particular module or several modules or entireprogram.

2. Determination of the goals of the test – to identify unauthorised/incomplete/inaccurate ineffective/inefficient code.

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3. Selection of testing approach – black-box or white-box.

4. Development of test data to accomplish the goal of the test.

5. Conducting the test through execution of test data or simulation of program execution pattern.

6. Evaluation of the test results against expected results and determination of the nature of discrepancies.

7. Documentation of the testing process.

Auditors should pay particular attention to the testing phase of the program development life cycle. Ahigh quality testing undermines how well other phases are performed. In a software project, testingconsumes considerable portion of development and implementation resources.

11.6.1 Levels of Testing

There are three levels of testing that can be conducted for testing of a program viz. unit testing, integrationtesting and whole-of-program testing.

11.6.1.1 Unit Testing

Unit testing refers to evaluating individual modules within a program. It is applicable for large programsin which each module constitutes substantive part of work, no matter whether the program is developedin-house or acquired. In the case of purchase of off-the-shelf program, this may not be applied.

Unit testing can be of two types viz. static analysis test and dynamic analysis test.

Static analysis test refers to evaluation of the quality of a module by examining the source code in thefollowing manner:

Type of test Detailment

Desk checking It is performed by the programmer who coded the module or bysomeone else by examining the module’s code for detection of syntaxerrors/logic errors/deviation from standards/fraudulent code.

Structured Walk Through The programmer who designed and coded the module leads otherprogrammers through the module to review the process for detectingerrors and irregularities.

Design and code inspection A special review team led by a trained moderator to review the codeof the program module in a formal manner. Documentation ismaintained e.g. checklists, results in preprinted forms, follow-upprocedures to ensure rectification of errors and irregularities etc.

A dynamic analysis test can be of two types:


Blackbox test Test cases are designed on the basis of requirements specificationfor the module and then executed to determine deviations fromrequirements. Internal logic of a module is not examined neitherexcesses to the specified requirements are determined.

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White-box test After the internal logic of a module has been examined, test casesare designed on the internal working of the module to traverse thedifferent execution paths built into a program.

Auditors should evaluate how well the unit testing was conducted. They may observe performing of staticand dynamic analysis by choosing a sample of test documentation.

11.6.1.2 Integration Testing

Integration testing refers to evaluation of groups of program modules to determine whether:

» interfaces are working properly,

» specified requirements are met,

» there is any degeneration under high workloads, and

» processing is carried out efficiently

There may be two different strategies viz. big-bang testing where all inter-module dependencies are testedtogether and incremental testing where subset of modules are tested in an iterative manner to reach thetotal group of program modules.

There may be three types of integration testing approach:


Top–down test Top level modules are tested first by simulating lower-level dummymodules to confirm the working of the interface correctly.

Bottom-up test Bottom level modules are tested first by simulating higher-leveldummy modules to confirm correctness of the working of theinterface.

Hybrid test It is a combination of top-down and bottom-up test. This is alsoknown as sandwich testing.

Auditor should gather evidence of integration testing and check whether systematic approach was adoptedand execution was carried out properly.

11.6.1.3 Whole-of-program Testing

Whole-of-program testing refers to test of the program in total to determine whether it meets the requirementand also evaluate the quality of the program.

Four types of tests are envisaged as under —


Function test Whether the integrated program meets the requirement. The test iscarried out by the programmers responsible for development.

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Performance test Whether performance criteria are fulfilled e.g. security, response time,throughput etc. The test is carried out by the programmers responsiblefor development of the program.

Acceptance test Carried out by the end users. The test comprises both function testand performance test.

Installation test Performed in the operational environment i.e. on the machine to beused for operational purpose.

While unit testing and integration testing may not be applied for smaller programs or off-the-shelf programs,whole-of-program test is applicable for all types of programs and of all sizes, bigger or small. Further,whole-of-program test may be carried out by the information system professionals or the end users or byanyone in the organisation responsible for development or implementation or acquisition of software. Anauditor should gather information and satisfy himself that the whole-of-program testing was designedand executed properly.

11.7 Outsourcing

Outsourcing may be defined as the use of parties, external to the organisation, to provide goods or servicesto the organisation. Outsourcing in today’s economic environment is considered as an important means ofimproving an organisation’s competitiveness and profitability. By outsourcing an organisation can paymore attention to its core competitiveness and take advantage of other organisation’s core competencies.

Any organisational function may be a point of consideration for outsourcing e.g. legal services, security,any manufacturing process, supply of components for the products, and even the information systemitself.

The reasons for outsourcing of information system function are:

1. exercising greater control over the information system function if it is outsourced e.g. vendors beingmore responsive to user needs, sharing economies of scale achieved by the vendor and restrictedconsumption of information system resources which are no longer free.

2. Innovative approach because of access to new technology and expertise expected from the vendor.

There is some preparatory work before a decision on outsourcing of information system services can betaken. Such work is of the following nature:

1. It should be determined what part of information system activities can be outsourced, what are strategicto the organisation, and whether suitable (expert, dependable, financially viable and reliable) vendorsexist.

2. Terms and conditions of the contract including termination clause for outsourcing should bedetermined incorporating scope of work, audit rights, performance criteria, responsibilities etc.

3. Monitoring compliance with the terms and conditions.

4. Impact on the organisation.

5. Procedures for outsourcing disaster recovery control.

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11.8 Internal Auditors vs. External Auditors

There is a controversy on the question whether the information system auditors should participate duringthe system development process. Reasons for prevention of systems development deficiencies at the outsetare weighed against the reasons for independence of the auditors to have review during ex post audit forbetterment. Three types of audit can be envisaged in respect of system development process:

1. Concurrent audit as members of the system development team.

2. Post implementation audit.

3. General audit for evaluation of overall controls, system effectiveness and efficiency.

External audit is generally considered in the area of general audit rather than concurrent or postimplementation audit. However, deployment of external auditor may be considered for concurrent and/or post implementation audit if it is opined that their involvement and expertise will be cost-effective andbring about quality in the process. Both internal and external auditors may have preference to preservetheir independence. If they participate as member of the system development team, they should not conductex post review, which, in effect, is evaluating their own work.

11.9 Audit Charter

There is a need for an audit charter for properly organising the information system audit function withinan organisation. An audit charter covers legitimacy and role of the internal audit function inclusive ofinformation system audit function and deals with three issues viz.

1. Place of audit function within the organisation and its role in contributing towards fulfilment oforganisational goals.

2. Authority of the audit function to gain access to records, facilities and personnel including Board ofDirector’s audit committee or to the Board itself.

3. Responsibility of the audit function to advice the management about the quality of attainment of thefour objectives viz. asset safeguarding, data integrity, system effectiveness and system efficiency.

The rights and responsibilities of both the internal audit function and the information systems audit functionwithin an organisation should be clearly defined to prevent disputes over the issue and ensure effectivenessand efficiency of the audit function. It is also to be determined whether information system audit shouldhave line function or staff function after considering the arguments for each strategy in the overallorganisational perspective. Another issue concerns about centralisation as against decentralisation. Thisissue is to be dealt with also in the case of external auditors. Much depends on whether the organisationalfunction is centralised or decentralised.

11.10 Information System Audit and Control Association (ISACA)

If an occupational group desires to be identified as a profession, it should satisfy five basic conditions asunder:

1. Existence of a common body of knowledge,

2. Existence of competency standards,

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3. Conduct of a valid examination for assessment of competency,

4. Existence of a code of ethics, and

5. Enforcement mechanism of the code of ethics and disciplinary measures.

In 1969, EDP Auditors Association (EDPAA) was founded in Los Angeles as a professional body. In 1994EDPAA changed its name to the Information System Audit and Control Association (ISACA). TheAssociation has more than 14,000 members spread over in more than 100 countries with local chapters inmore than 50 countries. It has a multi-volume publication called ‘COBIT’ (Control Objectives for Informationand Related Technology) where the focus is on business processes. There is a list of control objectives,controls to be used to achieve the objectives, and audit procedures to determine the effectiveness andefficiency of the control system. The local chapters conduct regular technical sessions for its members.ISACA has an active research and publication program. It also publishes a quarterly journal called ISAudit & Control Journal. It has developed model undergraduate and postgraduate curricula for informationsystems auditing.

11.11 Practical Examples

11.11.1 Guidelines to conduct audit of computerised accounting system

The guidelines to conduct audit of computerised accounting system will comprise the following steps:–

» Understanding basic features of the computer-based system

• input documentation

• accounting system

• management information system

• control system

» Review of application documentation

• data contents

• procedures

• internal controls

• system flow charts

• input formats

• record layouts

• coding system

• exception

» Evaluation of controls

• preparation of control checklists

• appraisal of key activities

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» Designing test procedures

• choosing testing techniques

• testing processing operations

» Using computers as audit tools

• test deck

• generalised audit program

» Preparation of audit program

» Compliance tests

» Evaluation of audit results

» Reporting to management.

11.11.2 A training program for the Internal Audit Department

A training program for the officers and staff of the Internal Audit Department may cover the followingtopics:

I. Introduction

» Overview of the company.

» Operations in various units and offices with future plan.

» Current information technology.

» Financial accounting and costing system followed.

» Present audit coverage.

II. New Computerised System

» Hardware, system software, file structure and processing techniques with special reference todistributed databases and client-server architecture.

» Information system designed.

» Codification.

» Input-output documentation.

» Data communication system.

» Decision support system.

III. Control And Audit Aspect Of Computer-based System

» Overview.

» Changes in accounting and auditing environment brought about by computers.

» Control system

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• Input controls

• Processing controls

• Output controls

• Internal controls

• Audit trail

• Security controls

IV. Auditing In Computer Based Data Processing System

» Introduction.

» Understanding basic features of the computer-based system.

» Review of Application Documentation.

» Evaluation of controls.

» Designing test procedures.

» Application of selected audit procedures to critical controls.

» Analysis of audit results and reporting.

11.12.3 Steps in auditing computer application systems

The steps in auditing computer application systems are as under :–

I. Familiarisation with the application system

» Basic features of the system

» Codification system

» Data validation, error handling and control system

» Accounting implications

» Audit trail

II. Scrutiny of application documentation

» Input-output formats

» Data contents

» Procedures

» Internal controls

» User and operation manual

» System flow charts

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III. Evaluation of system controls

» Checklists

» Preventive controls

» Detective controls

» Corrective controls

» Security controls including application backup and recovery procedures

IV. Designing test procedures

» selecting the testing techniques

» auditing around the computer

» auditing through the computer

» using computer as audit tools (test decks/generalised audit program)

» preparation of audit program

V. Compliance test of ‘key’ internal controls by application of selected audit procedures

VI. Analysis of audit results and reporting

» evaluation of the application under audit

» identification of weakness in the internal control system and impact thereof

» recommendation for improvement

» actions needed for implementation.

11.11.4 Example of outsourcing of internal audit jobs

Case Study

The Internal Audit Department of a public sector undertaking with diversified lines of production is centrallylocated in the Corporate Office in Kolkata. The company has seven manufacturing units in and aroundKolkata and one unit in Bihar. It has liaison offices in New Delhi and Mumbai. It also maintains GuestHouse in New Delhi and Mumbai. The company maintains a database in the corporate office and onlinetransaction processing accessed by terminals at various locations under a communication network. Thereare 3 officers and 12 staff in the Internal Audit Dept. 1 officer and 2 staff will be retiring within one year’stime. As a policy, there will be no replenishment. It has been decided that existing 6 stock verifiers will bemaintained for perpetual audit. Pre-audit of terminal benefits, pay related matters, strategic areas, systemreview, special reports for the management and audit committee, audit of unit in Bihar and liaison offices,replies to government audit will be continued with Internal Audit Dept. It is assessed by the managementthat the strength after retirement should be sufficient for this audit load. The remaining jobs may beoutsourced in phases for which monitoring and control is to be exercised by the Internal Audit Dept. Thejobs are required to be broadly categorised as per the above plan and submitted to the management forfurther direction.

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Probable Solution

Separation of jobs into two categories

Category – I : Jobs to be done departmentallty

Pre-audit Pay fixation New appointments, promotion, pay revision, step-up,arrears payment, increment.

Terminal benefits Gratuity, Leave encashment, voluntary retirementscheme, enfacement of service folders

Cash Verification Cash at Main, and all Units Imprest cash includingaccounting thereof

Perpetual / Perpetual stock taking All activities/Projects/Sites/Units – including GRN/

Yearly SIR valuation, redundancy, Stock-out

( Material ) Year-end verification Physical verification/valuation on test check basis

Post audit Employee-related Medical reimbursement, Interest subsidy on housebuilding loan LTA, TA/DA Advances to employeesForeign Tour

Departments Guest Houses in New Delhi and MumbaiUnit in Bihar, Liaison Offices in New Delhi and Mumbai

Other Routine Tender opening

Jobs Monitoring and reporting on Category II jobs

Special (CAG, Audit CAG audit repliesCommittee, Regular report and special requirement for Audit

Management Committee

requirement) Special requirement of management

Administrative Purchase and Stores Analysis

Control of rejected/surplus/non-moving materials

Scrap sale – disposal procedures.

Compliance of delegation of power

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Category – II : Jobs which may be outsourced

Purchase Capital Purchase proposal, Asset Register, addition/deletion/sale

Other than Purchase proposal test-checking/timeliness of

capital purchase procurement action/vendor rating

Demurrage

Material Material Issues to Bank guarantee/indemnity

With job workers for bond/return of scrap & surplus materials/rejections/

Job execution of jobs reconciliation/delivery control/insurance/excise

Workers formalities if applicable.

Services / Departments Medical Dept.

Infra– Transport Dept.

Structural Workmen’s Canteen, Electricity Consumption

Shop Machines (including Machine utilisation/idle time

Resources welding machines)

Labour Labour utilisation/idle time

Material Handling Utilisation/breakdown/idle time

Equipments

Tools & Consumables Procurement/utilisation /consumption/scrapping

Payment Creditors and Advance/proforma/rejections/bank guarantees/

Contractors Letter of credit/Liquidated Damages/delivery control

Collection Debtors Billing/Advance/deductions/Liquidated Damages/rejections/short-supplies/bank guarantee/IndemnityBond/Letter of credit/outstanding

PAYROLL Control aspect Earnings/deductions/reconciliation/control

Insurance Coverage/premiums/renewal/claims/realisation

COMPUTER Information System Efficiency of internal control system

Audit Control of input/output documents/time schedules.Audit trails Security controls & contingency plansHandling of users’ requirement/complaintsAudit through test packs.

SYSTEM Review of systems Purchase/Import/Stores/Cost-booking& procedures

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11.12 Short Notes

11.12.1 Auditing around the computer and auditing through the computer

Auditing around the computer requires the following steps:

a) Selection of one or more critical output from the computer system.

b) Verification of the results exhibited by the output to ascertain correctness and completeness of thetransactions processed.

c) Audit trail to locate the original source of input for verification.

Auditing through the computer implies verification of the computerised system itself and its efficacy toproduce the correct and required output.

11.12.2 Principal methods of using computers as audit tools

Two principal methods of using computers as audit tools are:

Test Deck which is made up of dummy transaction data containing both valid and invalid conditions.These are used to test the effectiveness of programmed controls. The test deck can either be created by theauditor himself on the basis of review carried out or procured from reputed vendors marketing standardsoftware package for the purpose. The processed result can then be compared with the predeterminedresult to ensure that the programmes are working accurately.

Generalised Audit Programmes which may be procured from the computer vendors/software companiesfor using computers in carrying out certain audit tasks e.g.

• Search and retrieve

• Test data generation

• Performing arithmetic operations

• Sorting, merging, matching, and comparing

• Copying

• Summarising

• Parallel simulation

11.12.3 Categories of controls

Preventive controls are designed to guard against risks inherent in certain data processing operations.

Detective controls are intended to arrest the consequences of what may bypass preventive control.

Corrective controls are detective types of controls that can be considered as complete only when keysupplement corrective controls like formation of audit trails, installation of adequate backup andrecovery procedures, implementation of automated error detection and correction techniques andother similar types.

11.12.4 Audit Trail

Audit trail refers to a system of designing of an information system in a manner that the historic data andinformation at any processing stage may be traced to verify the origin, correctness, authenticity, flow anddestination including the stages of security procedures for establishment of integrity of data and information.

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Problems for Self Examination

Q. 1. What are the characteristics of an Information System ?

Q. 2. Describe how growth of information system is contributing to business development ?

Q. 3. Describe how information system help in strategic planning.

Q. 4. Describe the requirement for a good Information System Infrastructure.

Q. 5. Draw a picture of good System Organisation and describe the role of different functional specialists.

Q. 6. Write short notes on

(i) High Level Language

(ii) Assembly Language

(iii) Debugging

(iv) Application software

(v) Virus

(vi) Fourth Generation Language

Q. 7. Describe the steps involved in System Development Life Cycle.

Q. 8. What is Systems flow chart ? How does it help in system designing ?

Q. 9. Describe the steps involved in Program Development.

Q. 10. What are major consideration for implementation of a system ? Describe the steps involved inSystem Implementation.

Q. 11. What are the requirements for a smooth System Change Over.

Q. 12. What are the components of systems development cost ?

Q. 13. Describe different types of file organisations and give comparison of these different types.

Q. 14. What is meant by Database Management System (DBMS) ? What are the advantages of DBMS overthe conventional file management system ?

Q. 15. Draw a DBMS System Structure .

Q. 16. Describe different models of DBMS Structure.

Q. 17. What are the Code’s Rules for stable Relational DBMS ?

Q. 18. Describe different activities involved in Data processing. What are factors you will be consideringfor deciding on mode of data processing.

Q. 19. Differentiate between batch processing and distributed data processing.

Q. 20. Write short notes :

(i) Protocol

(ii) Network Software

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(iii) Centralised architecture

(iv) Client/Server architecture

(v) Local Area Network (LAN)

(vi) Wide Area Network (WAN)

Q. 21. What is meant by Network Topology. Describe three different types of network topologies.

Q. 22. Differentiate between Internet and Intranet. What are common problems faced in Internet ?

Q. 23. Give the advantages of e-mail.

Q 24. How e-commerce can provide different types of prospect in business world ? According to you,what are the issues to be addressed for effective role of e-commerce.

Q. 25. What is Data warehousing ? What are the critical factors involved in data warehousing.

Q. 26. What is Data Mining ? Narrate the services provided by data mining.

Q. 27. What is meant by Knowledge Management ? How does it provide good solution in business ?

Q. 28. What is Business Process Re-engineering (BPR) ? Does it contribute in effective resource management.

Q. 29. Write short notes on :

(i) Expert System

(ii) Artificial Intelligence

(iii) Decision making process

(iv) Decision Support System

Q. 30. What are the objectives of developing a good Management Information System ?

Q. 31. What are the steps involved in developing an MIS ?

Q. 32. What are different approaches for developing MIS ?

Q. 33. What are the constraints commonly faced in operating an MIS?

Q. 34. Describe the limitations commonly encountered in operating an MIS ?

Q. 35. Why an ERP Package is getting popular these days ? How BPR helps in implementing an ERPSystem ?

Q. 36. What are the steps involved in implantation of an ERP System ?

Q. 37. What is the selection process for an ERP Package ? What are the cost components for implantationof ERP System ?

Q. 38. Name and describe some common modules of an ERP System ?

Q. 39. What are different risk factors in an Information system and how they are to be controlled ?

Q. 40. What methodology do you propose to minimise risk in operation of a system ?

Q. 41. What do you propose to create a proper processing environment ? What are the controls for a goodprocessing environment ?

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Q. 42. What are preventive measures do you propose against damage in processing ?

Q. 43. What is Computer fraud ?

Q. 44. What is the implication of auditing around the and auditing through the computer ?

Q. 45. Describe the various steps in auditing computer application systems.

Q. 46. Describe two principal methods of using computers as audit tools. On what basis do you categorisecontrols as preventive, detective and corrective?

Q. 47. Draft guidelines to conduct audit under the computerised environment.

Q. 48. Draw a training programme for the officers and staff of the Internal Audit Dept.

Q. 49. Discuss about the utility of documentation in supporting an information system. Is a programlibrary essential?

Q. 50. What are the criteria for recognition of a profession ? What is ISACA ?

Q. 51. Write short notes on:

(i) Audit Charter

(ii) Audit Trail

(iii) Outsourcing of Internal Audit Jobs

(iv) Evaluation of System Controls

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Revisional Questions


Question 1:

Write short notes on : 4×5=20

i) Open system

ii) lnternet

iii) Electronic Mail

iv) LAN, Ring Network, Star Network

v) Scanner

Question 2 :

(a) Briefly explain Decision Support Systems (DSS) illustrating how

they are used and by whom. 10

(b) What is an operating system ? What are its basic functions ? 10

Question 3 :

(a) What tools and techniques are used during system design ? 10

(b) What should be included in the written design specifications ? 10

Question 4 :

(a) What is an electronic mail ? 3

(b) What is modem ? Why is it required in data communication ? 8

(c) According to you what factors affect the security of the computer

system of an organisation ? 5

(d) What is meant by “backup” of computer files ? Why is it necessary to

keep back up of computer file ? 4

Question 5 :

Explain the following :

I. System concept 4×5=20

II. Source document

III. System & subsystem

IV. Debugging the programme

V. Data flow diagram

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Question 6 :

(a) Explain the concept of Data Base Management System. What are its benefits? 10

(b) Describe five input control techniques. 5

(c) What is the need for providing system control ? 5

Question 7 :

(a) What is system changeover ? What are the different methods of changeover? 10

(b) Explain the main stages of system project life cycle. 10

Question 8 :

(a) Distinguish between decision support system & executive information system. 10

(b) Explain briefly the main functions of an operating system. 5

(c) What is the difference between operating system software andapplication software. 5

Question 9 :

(a) What are the different types of file ? Explain the Transaction file. 10

(b) What is BUS and what are the different types of BUS. 10

Question 10 :

(a) What is Internet ? 10

(b) Write short notes on :

(i) DHTML

(ii) JAVA

(iii) e-mail

(iv) Extranet

(v) World Wide Web 2×5

Question 11 :

(a) What are the main objectives of MIS ? 10

(b) Explain the difference between LAN and WAN. 10

Question 12 :

Select True/False from the following statements : (1×15 = 15)

(i) All managers in a big organisation use the same type of information.

(ii) Sequential files are suited for on-line inquiry processing.

(iii) E-mail reaches its destination instantly.

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(iv) A computer is an electronic data processing machine used by several persons at a time.

(v) In LAN each computer can fulfil a function.

(vi) There is no difference between a LAN and a computer network.

(vii) CD-ROM are produced on a mass scale.

(viii) All hardwares features are available to an assembly language programmer.

(ix) Memory is used to store data, programs and results.

(x) CPU can read information directly from secondary memory.

(xi) Only executable files can be infected by virus.

(xii) Processing is done in the primary storage unit.

(xiii) Light pens and Joysticks are both pointing devices.

(xiv) To start an accessory, after positioning the pointer, double click the left mouse button.

(xv) On-line processing and real-time processing are same.

Question 13 :

Match words and phrases in left column with the nearest meaning in the right column.

(i) Arrangement of records in file (a) E-mail 1×10

(ii) Processing has fixed time constraints (b) File organisation

(iii) Transmission of letter, message, memos over (c) Operating systemcommunication net works.

(iv) Processing information of physical nature (d) Protocol

(v) Transferring programs from mainmemory to disk storage and back (e) Baud rate

(vi) Device dependent software (f) Icon

(vii) A program/utility under Windows (g) Swapping

(viii) Transmission speed of communication channel (h) Link

(ix) Any device to offer service to LAN users (i) Real-time processing

(x) Set of rules for orderly transfer of data among (j) Servernetwork users.

Question 14 :

(a) The HR department of a company is considering the implementation of a HumanResources Information System. They have two options-either to buy a ready-madepackage or to get it developed by a software services company. What are the pros andcons of each of these options? 10

(b) What do you understand by the term “system” ? What are its characteristics ? What doyou understand by integration of systems? 10

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Question 15 :

(a) What is an interface device ? Describe three such devices that are commonlyavailable. 6

(b) State the full form of —

(a) DSS

(b) MODEM

(c) SDLC

(d) EDI 4

(c) Describe briefly the basic components of a computer. If you wish to use yourcomputer for internet surfing and e-mailing, what would be the additionalrequirements in respect of hardware and software? Draw a diagram showing theconnectivity with the internet. 5+3+2=10

Question 16 :

(a) You as a Finance Manager, has received a proposal from your IT Department to enterinto an Annual Maintenance Contract (AMC) for an ERP software that has beenimplemented in the company. Considering the fact that ERP packages are provensoftware packages, why should the company enter into a maintenance contract? 8

(b) Write short notes on :

(i) Multiprocessing,

(ii) Distributed Data Processing 4+4=8

(c) State "True" or "False": 1×4=4

(i) A scanner is both an input and output device.

(ii) “Power Point” is not a spreadsheet software.

(iii) A computer virus and a program bug are not synonymous.

(iv) Data security and Data privacy mean the same thing.

Question 17 :

(a) “A business graph is a visual representation of numerical businessdata in an analytical manner so as to facilitate quick grasping of theinherent features.”

Name any five types of business graphs commonly used. 5

(b) In the context of internet, please define:

(i) E-commerce

(ii) E-money

(iii) E-mail

(iv) Webmaster

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(v) Yahoo !

(vi) Information highway.

Question 18 :

(a) “The success of a computer based information system greatly depends on a good codingsystem”. What is a code ? What are the characteristics of a good coding system ? Describeat least three methods of structuring codes with examples. 12

(b) Distinguish between : 4+4=8

(i) Master file and Transaction file;

(ii) Sorting and Merging of files.

Question 19 :

(a) How do you judge one CPU as more powerful than the other ? Name five major attributeswhich add up to making such decision. 5

(b) Match the following two columns :

(i) Search Engine (a) A method used to make hypertext documents readable on the World WideWeb.

(ii) HTML (b) A place where users can experiment with Internet services available, takeexpert assistance, undertake training courses, etc.

(iii) Web Browser (c) A list of common questions and sometimes with answers.(iv) FAQ (d) A Web site enabling users to access various levels of information.(v) HTTP (e) Web sites that allow users to communicate with various other users on-

line.(vi) JAVA (f) A program that enables a computer to download and view pages of the

Web.(vii) Chat Rooms

format (g) An encoding scheme used to create a Web document.(viii) Cybercafes (h) A language system that delivers programs to users which can then be run

on the users’ machines.(ix) Microsoft Excel (i) A programming language.(x) C++ (j) Spreadsheet package.

(c) Define a URL in the context of Internet. What role does it play ? 5

(d) What do you mean by ‘protocol’ ? What does it include ? 5

Question 20 :

(a) “Digital signatures do for electronic documents what handwritten documents do forprinted documents.” What is a digital signature ? How is it created andverified ? 10

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(b) State ‘True’ and ‘False’ and the basis for your decision : 2×5=10

(i) A System Designer ends his job where a System Analyst begins.

(ii) The higher the management, more structured are the problems.

(iii) ‘Data’ and ‘Information’ are synonymous.

(iv) In Binary Numbering System ‘bits’ and ‘bytes’ convey different meaning.

(v) Primary Storage and Secondary Storage are two sections of the CPU.

Question 21 :

(a) You have been engaged as Management Consultant for the computerisation planundertaken by a large manufacturing company. Before assessing the information needs,you have requested to identify at least three major activities under each of the followingfunctions of the company : 10

(i) Manufacturing;

(ii) Marketing; and

(iii) Finance and Accounting.

(b) Name three principal methods of creating databases. Describe any one method in detail.10

Question 22 :

(a) What is a knowledge-based system? How does it differ from an ExpertSystem? 10

(b) Explain the concept of database Management System. What is a Relational DatabaseSystem? 10

Question 23 :

A parcel carrying transport company has nine branches situated in South India. The top managementwant to have the figures on the various functions on a day to day basis, so that dailyperformance can be evaluated the next day.

(a) List out the data required for this purpose and give the formats for sending the datadaily by next day morning. 5

(b) How should these data be collected by the head office to evaluate the dailyperformance ? 5

(c) State the used of computers at branches and at head office. 5

(d) What precautionary measures should be taken to prevent the error ? 5

Question 24 :

(a) A company having multiple offices all over India, has decided to install an ERP packagefor its financial accounting in all its offices. Two options were considered :

(i) a centralized architecture in which the database will be stored centrally in HeadOffice and all offices will access it via VSAT connectivity, and

(ii) a decentralized architecture in which each office will maintain its own database,

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which would periodically be merged at Head Office for consolidation.What are the advantages and disadvantages of the two architectures? 5

(b) “While computerization has an array of advantages, it also exposes the company to anumber of hazards.” Name some of these hazards, and some measures to avoid them.

5

(c) Describe systematically the steps involved in program development. What do youunderstand by “top-down approach” to program design? 5+5

Question 25 :

(a) You have been given the responsibility of implementing a new software package inplace of a running old one. What techniques would you consider in changing overfrom the old to the new system ? 6

(b) What is Data processing? Describe the basic components of a computerized Dataprocessing system. 2+6

(c) What types of control are usually exercised on computer outputs ? 6

Question 26 :

Indulekha Fabricators Ltd. executes orders of general engineering nature. They have fourunits in and around Kolkata. They have recently introduced computer-based informationand accounting system, covering all the units including Corporate Office. They have alsodecided to introduce E-mail facility and maintain Web site of their own. As a result, changeshave taken place in the accounting and auditing environment, especially with regard to thecontrol system. The company has assigned you the task of preparing training materials fortraining the officers and staff of their Internal Audit Deptt. in order to be able to take up thisnew challenge. 8+8+4

Please prepare :

(i) the broad heading and subheading of topics to be covered in the training materials;

(ii) guidelines to conduct audit of computerised accounting system;

(iii) implication of auditing around the computer and auditing through the computer.

Question 27 :

(a) A company engaged in steel manufacturing activities is considering the implementationof an ERP system. The company has a few computerised applications running indifferent areas of the organisation. All these will be discontinueded after ERP system isimplemented. 15

A software firm has given a quotation for the new system which states that theimplementation will take a little more than a year and the capital cost will be Rs. 86lakhs (payable as Rs. 55 lakhs in the first year and Rs. 31 lakhs in the second year). Themanagement is wondering as to when the ERP system will recover all of its initial costsand start making a profit. What would be your answer based on the above data ?

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The following information about Operational Costs (Rs. lakhs) are also available :

Year 1 Year 2 Year 3 Year 4 Year 5

Old system 25 28 34 45 47New system — 7 14 15 16

(b) What do you understand by quality in regard to software ? Then what are tools usedfor quality assurance of software ?

(c) Write note on System Changeover Methods.

Question 28 :

(a) What is World Wide Web ? How would you distinguish it from Internet ? 2+3=5

(b) In a Purchase Order Processing system, a purchase order record has the following fields :

Field Name Maximum Field Size

Purchase-order-number 6Vendor Code 3Order-quantity 5Order-date 6

It is estimated that at any point of time, the outstanding Purchase Order filewould have a maximum of 750 outstanding purchase order records in the file(once material is received, the purchase order record is purged from the file).However, there may be a 15% increase in the total number of records in nearfuture. The file management software also requires an overhead of 20% forminimising probabilities of collision and overflow conditions. Compute the totalfile space requirements after allowing for 10% contingency factor on the total.7

(c) Distinguish between File Retention and File Recovery. 4

Question 29 :

(a) Distinguish between Data Integrity, Security and Privacy in regard to computerdatabases. 5

(b) Distinguisg between structured and unstructured decision. 5

(c) In a company, it is a regular procedure that the Sales Order master file on hard disk isbacked up on a Cartridge Tape at the end of each day. The system provides that all thetransactions that update the master file are available on a Transaction log file, which isalso backed up on a tape at the end of each day. During a particular day, both themaster file and the Transaction log file on the hard disk got destroyed. Draw a diagramshowing how the master file could be reconstructed on the disk with the help of backedup files. 10

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Question 30 :

(a) Your computer is considering three options to acquire software for computerising oneof its functional area. The options are :–

(i) Buying a readymade software package,

(ii) Engaging an outside software firm to develop the software, and

(iii) Developing the software in-house by own IT department. What are thepros and cons of each option ? 10

(b) What is prototyping approach of developing systems ? What are the advantages anddisadvantages of the same ? What are the two types of prototyping approaches to systemdevelopment ? 10

__________________

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Suggested Solutions to Revisional Questions

Answer to Question 1 :

i) Open System : An open system is one which interacts with its environment and can change itself toaccommodate the changes in factors like customers’ preference, price, product design etc. Theadoptability of an open system is judged by its capability in modifying the operational parameters ofthe system accordingly. It takes input from outside world and exports output to outside world. Forexample, a Financial Management system can function smoothly even with the change in creditperiod and credit limit from the banker is said to be having the flexibility to adjust itself with thechange the environmental parameters. If a system can accommodate all the changes in theenvironmental factors, as and when required, is said to be perfectly open system.

ii) Internet is worldwide network of databases and users. Any member of the lnternet can access any ofthe databases through his personal computer. Each user is given an ID number. The user can downloadtexts, graphs and data from other databases. Similarly the user can create database which can beaccessed by others. Internet facilitates international trade and commerce. lnternet helps in sharing ofknowledge and enables effective, cheap and fast communication across the world. It requires a modem,a personal computer, a telephone and the services of a service provider.

iii) Electronic Mail : This refers to transfer of information from one computer to another. The full expansionis electronic mail. It means that two computers, whatever may the distance between them, can beconnected together through communication technology. E-Mail enables one user to transferinformation, receive information and hold a dialogue with the other. The advantage is low cost highspeed communication.

iv) Local Area Network (LAN) is a network of computers within a limited geographical area, often withinthe same building but sometimes within a few kilometers as opposed to Wide Area Network (WAN)which can cover wide geographical area across cities and countries.

There are two configurations of computers in a network :

Ring Network - In a ring network the computers arranged in a circular fashion. There is no central or hostcomputer. The disadvantages of ring network is that the failure in any one computer in the network causesthe entire network to fail.

Star Network - In a star network, there is a host or central computer which is attached to the computers inthe network through multiple communication lines. All the computers communicate with each other throughthe host computer. Unlike in the ring network, the entire network does not fail if any of the computer in thenetwork fail.


a) Decision Support Systems (DSS) :

DSS is a combination of hardware and software to aid decision making. It includes Data Base (DBMS),Model Base (MBMS), Knowledge Base (KBMS) and Dialogue Systems (DGMS ). DSS is designed toaid the decision making functions of the Management. Management, at the lower levels, has to dealwith structured problems. Higher and Middle level managements, have to deal with strategic aspectsof decision making where the problems are unstructured. These levels of managements, need toolsthat can enhance their capabilities in the area of predicting the outcome of future course of actions. An

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important component of DSS is stimulation, i.e. construction of mathematical/statistical models forevaluating the behaviour of real variables.

Data Base Management Systems (DBMS) : This refers to a Data Base Management System softwarelike Ingress or Oracle, that helps in organising the data and facilitates easy accessibility through querylanguage for the decision makers, to retrieve any type of data depending on the decision environment.

Model Base Management System (MBMS) : This refers to a Model Base Management Systems thatfacilitates representation of a decision environment and stimulate the results of various alternativeplans of action. This helps in improving the quality of decision making.

Model base management systems include special purpose models like linear programming modelsetc. that help in decision making situations like production planning and optimisation.

Knowledge Base Management System ( KBMS ) : KBMS refers to Knowledge Base ManagementSystems. This is a recent development in applications of artificial intelligence. These systems capturethe experience of a decision maker in the form of rules so that rule base can be used by anotherdecision maker for decision making. These are known as Expert systems for Decision making.

b) Operating System :

An operating system is an integrated act of specialised programme that are used to manage the overallresources of and operations of the computer. It is specialised software that controls/monitors theexecution of all other programs that reside in the computer, including application program and othersystem software. Major functions of an operating system are —

i. Job Management

ii. Scheduling

iii. Acceptance of tasks

iv. Communication with terminals

v. Computer resources management

vi. Data management

vii. Job accounting

viii. Access control.


(a) The following tools and techniques are among those used during system design —

(i) Organisation standards – Some organisations have standards/manuals that specify a consistentdesign approach. The procedures to follow in designing output reports, input forms, andprocessing logic may also be spelled out.

(ii) Top-down design methodology – The top-down design techniques requires the early identificationof the top-level functions in the proposed system. Each function is then broken down into ahierarchy of understandable lower level modules and components. A top-level chart shows howthe total structure of the system is prepared, lower-level diagrams are created to show the input/

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processing/output details of each function, module, and component. Several iterations will usuallyoccur in this design and charting process. Designers may start with simple diagrams showinggeneral solutions. This first effort is then refined to produce more complete charts as the designrequirements become clearer.

(iii) Design reviews and walkthrough – Periodic sessions may be held so that interested users canreview the design process. Users can be encouraged to look for errors and to make commentsduring this design walkthrough.

(iv) Special charts and forms – The charts and forms prepared during the system analysis phase arevery helpful in the design stage. Weaknesses spotted during analysis can now be corrected. Andthe existing input/processing/output relationships may now be used to design a more integratedsystem.

(v) Automated design approaches – Special computer programs can be used to evaluate varioushardware/software alternatives during the design stage. Processing requirements can be supplied,and a computer program can then determine how efficiently these requirements can be processedusing different equipment alternatives.

(b) Usually the designers have prepared a detailed set of written and documented system specificationsto achieve the system study goals. These specifications should include :–

(i) Output requirements – The form, content, and frequency of output is needed.

(ii) Input requirements – The necessary new input data should be identified along with the storedfile data that are required.

(iii) File and storage requirements – The size, contents, storage media, record formats, accessrestrictions, and degree of permanency of any affected files should be known.

(iv) Processing specifications – The procedures needed for the computer to convert input data intodesired output results should be indicated. Manual processing procedures should also be noted.

(v) Control provisions – The steps required to achieve system control should be specified, and thelater system testing and implementation procedures should be outlined.

(vi) Cost estimates – preliminary estimates of the costs of different alternatives should be made.


(a) E-mail is the single most important reason to use the internet. It is quite similar to the postal mailingsystem. The service is provided by E-mail service provider, who maintains a computer with a largestorage place. The customers are provided an address, which is used for communication. A customerdesiring to communicate with another, sends the mail to the address of the receiver using a specialsoftware. If the computer of the receiver is not in use, the message gets stored in the service provider’scomputer. When the receiver switches on his computer, he is immediately informed that he has a‘‘mail’’, he can read the mail and answer it. The advantage of E-mail is that the message get deliveredeven when the receiver is not available. But the system, in addition to a telephone needs a computerwith proper software and a modem.

(b) A Modem is an electronic device which is used converting digital signals to analog waves and vice

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versa. The process of converting a digital signal to analog signal is called modulation and reverseprocess is called demodulation. Modem is a short form for Modulator - Demodulator. Modem isessential for data communication. If the telephone system is to be used for connecting up with distantcomputers, as the signal generated in a computer is digital and those handled by the telephone system,so for is analog.

(c) Factors affect the security of the computer system - Normally there are three types of factors whichmay affect the security of computer system of an organisation —

Physical factor

(i) Non–protection of computer hardware from fire, flood, theft, dust, humidity, earthquakes, powerfluctuations etc.

(ii) Non–protection of computer software from fire, flood, theft, dust, humidity, earth quakes, powerfluctuations etc.

(iii) Non–protection of building housing with the computer facility from fire, flood, earthquakes, etc.

System factor

(i) Loss of data, software, damage to equipments.

(ii) Malfunctioning of the system due to hardware failure and error in software.

(iii) Non-availability of backup of data/Software.

Human factor

Risk due to unintentional/intentional activities of others. Unintentional activities may be due to accidentaleffects of wrong changes to data, wrong deletion of files/programs, running of a wrong program, forgettingthe password etc. Intentional activities are unauthorised deletion/corruption of data/files etc.

(d) The storage devices used, especially the disks, in a computer system are not very reliable beingsusceptible to becoming bad, which may be caused by "bad sectors", the storage location becomingpartially bad. Also, the possibility of total disk failure, called ‘‘erasing’’ can not be ruled out. All thesewould cause loss of the data/program stored. To avoid such a loss, a duplicate copy of the data andprogram files are kept in a separate storage devise, which is called back ups. For example, the filesfrom hard disks are backed up in floppies, tapes, or tape cartridges. A separate utility is used forcrating back ups, where the files which are being copied, are combined to make better utilisation ofthe storage space.


I. System concept : The term "system" is used often and in different ways. A system is a group ofintegrated parts that have the common purpose of achieving some objective(s). The key characteristicsof a system are —

• A group of parts - a system has more than one element

• Integrated parts - A logical relationship must exist between the parts of a system.

• Common purpose of achieving some objective(s). The system is designed to accomplish one ormore objectives. All system elements should be controlled so that the system goal is achieved.

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II. Source document : Data must be originated in some form and verified for accuracy prior to furtherprocessing. They may initially be recorded on proper source documents and then converted into amachine - usable form for processing or they may be captured directly in a paperless machine-readableform.

III. System and subsystem : A Computer is a group of integrated parts that have the common purpose ofperforming the operations called for in the program being executed. It also is a system. Any systemmay be comprised of smaller systems or sub-systems. A subsystem is a smaller system containedwithin a larger one.

IV. Debugging the program : The first step in program implementation is to debug the program i.e. todetect and correct errors that prevent the program from running. Testing the results produced by theprogram to see if they are correct is the next implementation step. And ensuring that a completedocumentation package is available for the application is a third implementation step.

V. DFD is a simple method of refreshening input, process and outputs, within an information systemand can be suitably adapted for use at the data gathering stage of system analysis, to depict proceduresas they are being currently carried out, as well as, for showing the flow of data through the programat design stage along with transformations in data, resulting from processing.


a) Concept of DBMS - Data Base Management System (DBMS) is a tool for managing information as a'resource which overcomes the limitations of traditional file design. Its main focus is one ease of filecreation, reorganisation summary, report generation and updation. In short it aims at easing thetransaction processing routine and information access. It seeks to avoid duplicate data creation.Information is the resource of a Company. Hence it is essential that the discipline and importancetraditionally apply to the management to the other resources like man, machine, materials and moneyare applied to information too. Database management systems manage the data of a company in sucha way that the data are structured and related, and hence obtainable in a more orderly and logicalfashion.

In traditional file design, files are designed for a specific reporting requirement with little regard tofiles designed for other reports. For example, one analyst designed a customer master file for anaccounts receivable system without any regard to the customer file designed in a related system likesales order processing which also maintain data about customers. As a result, there were inconsistenciesand redundancies of data among various information systems. In the DBMS environment there aretwo individual files for a particular system. Rather, a centralised data base is created and controlledwhich becomes common to all information systems.

Generally DBMS performs the following functions :

• Organises data in a manner most suitable to each application

• Integrates data by establishing relationship between data elements.

• Separates data by distinguishing between logical description and relationship of data from theway the data is physically stored.

• Controls how and when data is physically stored

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• Locates and provides data to the programs

• Protects data from unauthorised accesses, operating system failures, inconsistencies in updatesetc.

Benefits of DBMS :

• Reduced programming costs because many input-output routine normally coded by programmersare handled by DBMS itself.

• Reduced system maintenance cost because of many common causes of system failures like input/ output, file description, etc. are handled via DBMS.

• Reduced data redundancy because each element of data is maintained by a single source.

• Increased reliability of data because of checks and control mechanisms inherent in DBMS.

• Consistency of data because of each element of data is maintained in one place only.

• Easy retrieval of data through query language.

b) Input Control Techniques :

1. Editing : Checking for completeness, validity and consistency of data.

2. Item counts and control totals : In a batch processing system, the number of items being inputand total of certain fields. For example, number of invoices being entered and the grand total ofall invoices.

3. Sequential number of documents : Numbering all input documents sequentially and checkingwith the total of documents entered.

4. User codes and password : Only those people having a user code and password will have accessto the system and be able to input data.

5. Recording of receipts of documents : Recording of logging in the data, volume and otherinformation about data received from outside the system.

c) Needs for providing system controls

• internal control is a significant management obligation. System control is a part of the internalcontrol mechanism.

• With increasing computerisation, the traditional security, audit and control mechanisms take ona new and different form.

• The increasing complexity of complex on-line data communication oriented system makecompanies vulnerable to unauthorised manipulations of data, program and operating procedures.

• Many of the computerised systems now cut across many departments, lines of responsibilityand even geographical boundaries. Hence the traditional control mechanisms become ineffectivein many cases.

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a) System Changeover :

The changeover is the process of changing over from the old system to new system. It is the last stageof system development process. When the new system has been developed, it has to replace the oldsystem. The process begins when the programming and testing of the new system has been completedand by training the users with the new system, by creating the new data base and by installing thenecessary equipment. It is a critical phase in that the method and style of changeover will determinethe users response and attitude towards the system.

Different Methods of changeover :

Three alternatives are available for changeover from the old system to the new system -

• Crash Method - The old system is discontinued and the new system is implementedsimultaneously.

• Parallel method - The old and the new system is run concurrently for a period until the newsystem works to the satisfaction of all concerned.

• Staged method - The old system is replaced gradually by stages.

b) The total process of developing a system under the modern approach is divided into four stagescalled —

(1) System Analysis

(2) System Design

(3) System Implementation

(4) System Support

In the first phase, it can be termed as the starting phase, an analysis is made of the existing system ofworking to draw an outline of the new system which would meet the requirement of the user. Thesecond stage can be termed as the development phase, when the detailed design of the proposedsystem is drawn up. It also involves acquiring the hardware with necessary supporting software. Thethird phase called implementation stage, involves implementing the new system designed, i.e. makingit operational. The last stage called system support phase is involved in giving finishing touch to thesystem developed. However, it should be noted that these four stages are not isolated compartmentsand these are also not followed serially - often one has to go back to the previous stage to incorporatemodifications and new ideas, till the total job of successfully commissioning the system is completed.


(a) Decision Support System (DSS) are programs tailored to the management needs and designed to aiddecision making functions. The lower level management normally deals the structured problem whichinvolves processing of transactions, keeping records and providing reports or routine activities. Higherand middle level management deal with strategic and technical aspects of business decision. Theytherefore need aids which enhance their capabilities in the area of predicting the outcome of future

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course of actions. An important component of DSS is stimulation i.e. construction of mathematical/statistical models for evaluating the behaviour of real variables. DSS includes DBMS, KBMS anddialogue system (DGMS).

An Executive information system (EIS) is a refinement of the DSS which enables an executive tonavigate through the data base and view the database from multidimensional angles.

An EIS is a software that can facilitate navigation in any dimension within the database created out oftransaction processing systems.

(b) The main functions of any operating system (OS) are as follows –

1. Processor Management : The OS assigns different tasks to the processor and monitors theirexecution.

2. Memory Management : The OS allocates main memories and other storage areas to the systemprograms and other programs and data.

3. Input / Output management : It coordinates and controls the working of the various input/output devices like the Disk Drives, the Monitor, the Printer etc.

4. File Management : It handles the storage of information in the form of files on various storagedevices and transfer of these files from one device to another.

5. Interpretation of commands : The OS interprets the commands and instructions issued by theuser or an application program and executes them.

6. Job priority : It determines and maintains the order in which jobs are to be executed in thecomputer system.

(c) Operating system software : The operating system software is a set of program instruction that act asa interface or layer between the user and the application program and the computer hard ware. Itsupervises and directs the operations of the computer. Essentially, it performs three tasks : it managesdevices like storage and retrieval of files from the hard disk, it controls program execution and processescommands that are entered by the user. Some popular operating software are MS-DOS, UNIX,NETWARE, WINDOWS-95 etc.

Application software : An application program is one which is designed to be used for a specificpurpose that is useful to man. Word processing, Accounting are some specific application areas forwhich such software is written. All Transaction processing is done through Application programswhich can vary from simple accounting packages to complex Enterprises Resource Planning Packageswhich uses a Database package as backend.

(d) ROMs and RAMs : Semiconductor memories are of two types : RAM (Random Access Memory) andROM (Read Only Memory). RAM is a read / write memory. Information can be written into and readfrom a RAM. It is a volatile memory. It stores information so long as power supply is on. When powersupply goes off or interrupted the stored information in the RAM is lost. ROM is a permanent typememory. Its contents are decided by the manufacturer and written at the time of manufacture. RAMsof various capacities are available, for example, 1M, 4M, and 16M etc.

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(a) There are five broad categories of Data file used in any information system.

Master

Transaction

Work

Backup

Security

Audit

Transaction Files

Transaction Files are files in which the data relating to day to day business events are recorded, prior to afurther stage of processing. This further processing may be the use of the transaction data to update masterfiles, or the arriving of the transaction for audit purposes. After the transaction file is processed it usuallyre-initialised, and further transactions are recorded in it. Examples of transaction files are :

Customer’s Orders for products (to update an order file)

Details of price changes for products (to update a product file)

Details of cash posting to customer accounts (to be held for audit purposes)

Once a master file has been created, it must be kept current. File changes come from a number of sources,including business activities, the activities of individuals or changes in governmental policy. The documentsor data created by these activities are commonly called transactions.

Transaction data may be collected automatically or may initially be recorded on source document andlater converted to a machine-readable format.

(b) The BUS

The various chips on the motherboard have to be connected in some way, so that they can pass signalsback and forth and communicate with each other. This is done by setting up a common communicationchannel i.e. a set of wires that acts as a common carrier for signals passing from one component toanother. This channel of wires is called a bus. These buses also provide an electrical interconnectionbetween the processor components and the interface devices used with peripheral equipment (serialinterface, parallel interface).

There are three different types of buses on the motherboard which connect all the components witheach other. These are : The Data Bus, The Address Bus and The Control Bus. All instructions consist oftwo parts, the first part of the instruction indicates the operation to be performed and the later partcontains the address of the data that are to be used by the instruction (not the data itself). The processorsends the addresses over the address lines of the bus to RAM. From the RAM, it gets back the contentsof that address over the data lines of the bus. The processor then executes the instruction. The Controlbus carries the control signals to and from the control sections of the processor thereby controlling theentire processing.

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(a) The Internet is a worldwide network of networks. It is a conglomeration of smaller networks andother connected machines spanning the entire globe. It consists of over thirty thousand networks in71 countries. Around the world, each country has at least one backbone network that operates at avery high speed and carries the bulk of the traffic. Other smaller networks connect to that backbone.

The Internet has no central authority – networks on the Internet simply agree to cooperate with eachother. Individual computers on this networks can share files and transmit information. Variousnetworks connected to the Internet format messages with Internet Protocol (IP). IP is often used inconjunction with TCP (Transmission Control Protocol). The combination of TCP/IP have made theInternet robust and flexible.

(b) (i) DHTML (Dynamic Hyper Text Markup Language) : This is the latest version of HTMLthat is still being designed. DHTML solves a huge list of problems associated with laying outwebpage designs.

(ii) JAVA : This is a new language. Destined to change the Internet in a way very wonderful. Thislanguage can allow you to do incredibly versatile things via web pages running programs.

(iii) e-mail : Electronic mail - It is an electronic means of sending message from one computer toanother in an organised fashion. Sending e-mail uses the SMTP protocol

(iv) Extranet : An extension of a corporate internet. It connects the internal network of one companywith the internets of its customers and suppliers. This make it possible to create e-commerceapplications that link all aspects of a business relationship, from ordering to payment.

(v) World Wide Web : It is also known as just “the web” or WWW , this is perhaps the main reasonfor the internet’s growing popularity in recent years. Based on a client server architecture, theweb consists of numerous servers on the internet. A server is identified by an address in a specialformat, called the Uniform Resource Locator.

A web site is a computer server with databases and connectivity. It has homepages of its own or for itscustomers on rent. A homepage can be set up at customer’s own site with appropriate information.The homepage can serve as a mirror of an organisation that provides financial data, on-line orderforms etc.


(a) A Management Information System (MIS) should provide –

(i) Current information for monitoring and control.

(ii) Accurate information about the system

(iii) Timely information out of large volume of data.

(iv) Economical information (low cost )

(v) Relevant information for decision making.

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(b) Decision Support Systems (DSS), is a combination of hardware and software to aid decision making.It includes Database (DBMS), Model Base (MBMS), Knowledge base (KBMS) and Dialogue Systems(DGMS). DSS is designed to aid the decision making functions of the management. Management atthe lower levels has to deal with structured problems. Higher and middle level managements have todeal with strategic aspects of decision making where the problems are unstructured. These levels ofmanagements need tools that can enhance their capabilities in the area of predicting the outcome offuture course actions. An important component of DSS is simulation, i.e. construction of mathematical/statistical models for evaluating the behaviour of real variables.

Data Base Management System (DBMS) : This refers to a database management system software likeIngress or Oracle, that helps in organising the data and facilitates easy accessibility through querylanguage for the decision maker, to retrieve any type of data depending on the decision environment.

Model Base Management System (MBMS) : This refers to Model Base Management System thatfacilitates representation of a decision environment and stimulates the results of various plans ofactions. This helps in improving the quality of decision making.

Model Base Management System includes special purpose models like linear programming modelsetc. that help in decision making situations like production planning and optimisation.

Knowledge Base Management System (KBMS) : KBMS refers to Knowledge Base Management System.This is a recent development in applications of artificial intelligence. These systems capture theexperience of a decision maker in the form of rule so that such rule base can be used by anotherdecision maker for decision making. These are known as expert systems for decision making.

(c) Difference between LAN and WAN :

1. A LAN is restricted to a limited geographical coverage of a few kilometres, but a WAN spangreater distance and may operate nationwide or even worldwide.

2. The cost to transmit data in a LAN is negligible since the transmission medium is usually ownedby user organisation. However in case of WAN, this cost may be very high because thetransmission medium used are leased lines or public systems such as telephone lines, microwaveand satellite links.

3 In a LAN the computer, terminals and peripheral devices are usually physically connected withwires and coaxial cables. Whereas in a WAN there may not be a direct physical connection betweenvarious computers.

4 Data transmission speed is much higher in LAN than in a WAN.

5. Fewer data transmission errors occur in case of a LAN as compared to a WAN. This is mainlybecause in case of a LAN, the distance covered by the data is negligible as compared to a WAN.


(i) F

(ii) F

(iii) F

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(iv) T

(v) T

(vi) F

(vii) T

(viii) T

(ix) T

(x) F

(xi) F

(xii) F

(xiii) T

(xiv) T

(xv) F


(i) B

(ii) I

(iii) A

(iv) H

(v) G

(vi) C

(vii) F

(viii) E

(ix) J

(x) D


(a)

Type of Software Advantages Disadvantages

Buying 1. Can be seen and tested 1. May not meet requirements fully.

readymade 2. Will have few bugs. 2. Enhancements / modifications

3. May be less costly. may be expensive & time consuming.

4. Time saving in implementation. 3. Procedures, methods, Forms, etc

5. Documentation/help facilities. may have to be changed.

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Type of Software Advantages Disadvantages

Built by an 1. Better control over schedule 1. Difficult to negotiate on effortoutside firm and cost and time required.

2. Does not affect day-to-day 2. Difficult to implement withoutsystems operations due to adequate technical knowledgedevelopment work. of the software

3. Ready skills when new 3. Difficult to maintain afterhardware or software implementation.platforms.

4. Unfamiliarity of the outsideparty with business mayhamper quality.

(b) A system is an organised grouping of components-persons, methods, machines and materials thatare collectively set to accomplish some specified objectives.

The characteristics of a system include Objectives, Components, Structures, Behaviour and Life Cycle.

When all information in an organization is channeled into a common data base to serve the informationrequirements throughout the organisation. It is termed as integrated System.


(a) An interface device is a device for communication between computer and human beings. Three suchdevices are

Mouse : This is the most common interface device on PCs. It has a rubber ball at the bottom and twoor three switches at the top. When the mouse is moved on the desktop or on a pad, the pointer on thecomputer (mouse pointer) also moves on the screen of the computer. The rubber ball provides themovement of direction and a click on a switch on the mouse triggers action command to the computer.

Trackball : This is similar to the mouse. The difference is that unlike the mouse the trackball is fixedon a case on the computer (usually on the base of the computer). The trackball is more common inportable computers.

Joystick : With this, moving a lever on the base of computer carried out the movement of the pointeron the computer.

Cursor : It is a communication device between the user and the machine.

(b) DSS : Decision Support System,

MODEM : Modulator Demodulator

SDLC : System Development Life Cycle,

EDI : Electronic Data Interchange.

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(c) The basic components of a computer are –

» Input device ( e.g. Mouse, Keyboard etc.)

» Processing Unit consisting of Memory, Arithmetic / Logic Unit and Control Unit.

» Output device (e.g. printer).

For Internet surfing or e-mailing, the additional units that will be required are ( i) Modem (ii) TelephoneLine (c) Browser software like Internet explorer or Netscape Navigator. In addition, one will have tosubscribe to an Internet Service Provider (ISP).


(a) All software irrespective of how well it has been designed and developed and irrespective of whetherit is an ERP or any other packaged software requires maintenance over its life time. Maintenance ofinformation systems means enhancement modifications and corrections to the software. There areseveral reasons why all information systems need maintenance viz.

• Changing business environment.

• Changes in users' expectations.

• Changes in technology.

• Deterioration in software quality due to patches over time

• For time savings.

(b) (i) Multiprocessing is the term use to describe a processing approach in which two or moreindependent processors are linked together in a coordinated system. In such a system, instructionsfrom different and independent program can be processed at the same instant in time by differentprocessor. The processors may simultaneously execute different instruction from the sameprogram.

(ii) Distributed Data Processing (DDP) is a system in which the computing and other resources aredecentralized instead of being located at just one place. The points at which the resources arelocated are interconnected and although may act autonomously at times also at cooperatively inhandling common problem.

A DDP system therefore has a number of characteristics :

(i) Computer hardware and user are decentralised.

(ii) There is some form of communication linkage between location where the computer resourcesand user are situated.

(iii) Users tend to have responsibility for there own data and data processing - they tend to 'own'their system.

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(iv) When the system is not operating as a whole, quite often some data processing can continuelocally at individual user sites.

(c) (i) False , (ii) True , (iii) True , (iv) False.


(a) Five types of commonly used business graphs are :

• Column Chart.

• Bar Chart

• Line Chart

• Pie Chart

• Stacked bar Chart.

(b) (i) Electronic Commerce refers to the business by way of sale of goods and provision of servicestransacted electronically over the Internet. It is an on-line approach to conducting business withcustomers including advertising marketing order entry and processing, payment, and customersupport. The Internet is an open system and the users can hardly be traced. Important features ofe-business are :-

Five types of commonly used business graphs are :–

» Electronic signature.

» Cryptography

» Authentication and

» Certification.

(ii) E-money refers to the digital version of currency and exchange. Digital money moves through amultiplicity of networks rather than via current banking system. Electronic currency substitutesfor money on on-line transaction, including secured credit cards, electronic cheques and digitalcoins.

(iii) E-mail is system that enables a person to compose a message on a computer and transmit itthrough a computer network to another computer user.

(iv) Webmaster is the individual responsible for maintaining and updating the content of a WorldWide Web document. Webmasters are the creative force behind the World Wide Web.

(v) Yahoo ! It is one of the most popular website having lot of facilities like Browsing, E mail, Chatetc. It also have a search Engine

(vi) The Information highway refers to the interconnected series of networks that provides theinfrastructure for transporting information throughout the world.

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(a) A code consists of a number of digits to denote a specific entity. Thus a lengthy description may beexpressed by unique code, which facilitates designing input documentation, and computerisedprocessing. Even in a manual system, use of codes are useful in day-to-day transaction andcommunication e.g. work order no., sales invoice no., purchase order no., form no. etc. Employeesmay have identification no./ ticket no./ employee no. and selected few letters from the name. Thusemployee no. may be matched with the selected letters for validation process for the sake of accuracy.There are other methods of validation process also. Codes representing major key fields are matchedbetween the master file and the transaction file in a computerised environment for data processingpurpose.

A good coding system should have the following characteristics :

• Each code should be unique, compact and meaningful.

• Size of the code should be precise, and of fixed length with flexibility to facilitate insertion or expansionas may be envisaged.

• A code of a major key field may have a check digit embedded in it to ensure accuracy of processing.

• Codes should consist of numeric digits as much as possible to facilitate validation and faster processing.

The following three methods of structuring codes are more commonly used :

• Sequential Code : Codes are allotted as consecutive numbers to data entities e.g., employee number,invoice number, purchase order no., bill no. etc.

• Block Code : Under this method, entities are classified into groups and blocks of numbers are assignedto each group. The block size depends on number of items expected to be in use in the classifiedgroups with flexibility for future insertion/exception. For example, an organisation with 7000employees may have a 5-digit employee number where the left most digit may indicate category ofemployees viz., 1 for officers, 2 for supervisors, 3 for clerical staff and 4 to 9 for workmen. The rightmost digits may be reserved for check digit. Thus a block of 1000 numbers are reserved for each of thethree categories of officers, supervisors, and clerks and rest 6000 numbers for workmen (first 1000numbers beginning with left most ')' are left unallotted at present).

• Bar Code : Such codes are popular in companies running departmental stores and consist of verticalline of various thickness to represent digits 0-9. The code identifies the product, the manufacturer,and the price of the product on which it is affixed. The line symbols are easily read by the computerand converted into numbers that represent the code. One of the popular methods is UPC or UniversalProduct Code.

(b) A master file contains information of semipermanent nature e.g. employee number, name, address,date of birth, date of appointment, designation, scale of pay etc. Such a file is called employee masterfile. Insertion, deletion and modifications in the fields are carried out regularly to keep the master fileupdated. Payroll processing is dependent on both the master file and the transaction file accessedsimultaneously.

A transaction file is of temporary nature and contains records of current transactions that are variablein nature from period to period e.g. stores receipts and issues during a period. Such transaction file

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together with the master file are used for updation of master files and giving printouts for varioususers.

(c) Sorting constitutes arranging records in a file in either ascending or descending order of a key field.Merging involves combining two or more fields into a single file considering one or a combination ofkey fields as the sequence number. Sorting and merging is an essential element of data processingactivity.


(a) (i) Memory cycle time.

(ii) Storage Capacity

(iii) Number of data transfer channels

(iv) Range of multi-programming capability

(v) Real time processing capacity

(b) (i) – (d)

(ii) – (g)

(iii) – (f)

(iv) – (c)

(v) – (a)

(vi) – (h)

(vii) – (e)

(viii)– (b)

(ix) – (j)

(x) – (i)

(c) URL is the acronym of Uniform Resource Locator. It contains information about the location of adocument. When a user clicks on a link, the URL provides the information about the link to thebrowser. It assists in implementing any link from one document to another.

(d) Protocol refers to computer software that handles transmission between computers and allowscomputers to communicate. It is a set of standards defining the procedure and the format for data tobe transmitted via communication links.


(a) ‘Digital signature’ means authentication of any electronics record by a subscriber by means of anelectronic method or procedure.

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Digital signature is created by using a hash result which is unique to both the signed message and agiven private key. For the hash result to be created there must be only negligible possibility that thesame digital signature could be created by the combination of any other message or private key.

Digital signature is verified by the process of checking the digital signature by reference to the originalmessage and a given public key and determining whether the digital signature was created for thatsame message using the private key that corresponds to the referenced public key.

(b) (i) False. The System Analyst is concerned with analysing the existing system and preparingspecifications for the proposed system, which are passed on to the System Designer for detaileddesigning.

(ii) False. Generally lower level management deal with structured problems. Middle level is concernedwith tactical planning while top level management deal with strategic planning and controlhaving nature of unstructured or semi-structured problems.

(iii) False. Data pertain to raw facts that are assembled, analysed and summarised to produceinformation, which is meaningful for the recipient.

(iv) True. BIT is an abbreviation of Binary Digit consists of 0 and 1. Byte is a combination of 8 bits,which represents either an alphabet, digit or any special character.

(v) False. Primary or main storage is a section of the CPU while secondary storage holds dataoutside the CPU. The CPU handles the secondary storage like any other input-output peripheralsand can access data in the secondary storage only when the same are brought into the mainmemory. So, we can say that, the data stored within the secondary storage are dumb data andthat when transferred to the main memory become live data.


(a) (i) Manufacturing :

(i) An order schedule (to hold shipping schedules, manufacturing and deliveryschedules, for each order)

(ii) A sub-assembly pole (listing all intermediate assembles for final products)

(iii) A price list file (listing all component required by each assembly), and

(iv) An availability file (containing information about availability of all the required componentsand sub-assemblies).

(ii) Marketing :

(i) Order filing (including order entry shipping and billing)

(ii) Order getting (sales promotion, customer relations, advertising)

(iii) Information and analysis

(iv) Planning (using techniques such as forecasting, simulation etc.)

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(iii) Finance and Accounting :

(i) Logistics-related activities

(ii) Collection and dissemination of financial information

(iii) Analytical activities

(iv) Share accounting

(v) Budget and finance planning

(vi) Tax planning

(vii) Resource mobilisation

(b) Three principal methods of creating databases are :

(i) Hierarchical (or tree) databases : In this type of database, records are arranged in the constructionof an organisation chart, i.e., an inverted tree in which the top record may be viewed as the root,with the next level having several nodes (or child records), with the child record becoming aparent for the records (nodes) at the next level, and so on. When a parent record has no child, thelatter is termed as a leaf.

(ii) Network Database : This is also a kind of hierarchical database but with certain differences. Inthis type of database each node (or child) may have several parents. The nodes are interconnectedin a multidimensional manner by means pointers which are established by using one of thefields in each record to contain address of the physical medium of the next record in a logicalsequence. DBMS software makes it possible to reach any node for the purposes of data retrieval.

(iii) Relational Database : The relational structure is the most popular, (particularly with PCs), andis, possibly, the most flexible type of data organization for database construction. This type ofdatabase is built around the concept that data contained in the conventional files can, with someduplication, be transformed into two dimensional tables. Each such table may have any numberof rows, but may have only few columns. Data in each column has to be of uniform size, residinga field in a record, for example customers' name.

(c) Types of Databases by ownership criteria are :

(i) Individual databases

(ii) Company (or shared databases)

(iii) Distributed databases

(iv) Proprietary databases.


(a) Knowledge-based system is a software developed around facts and generally accepted rules forperforming specialized tasks. Human intelligence duplicated in such systems are of surface type andare not of very complex in nature.

An expert system, on the other hand, is also a knowledge -based system. But, it is developed aroundArtificial Intelligence to capture deep knowledge and expertise in particular field.

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(b) The term Database Management System or DBMS is applied to a specialised kind of software thatmakes it possible to create and modify a database and allows access to the data stored in the database.It essentially consists of two parts viz. Data Dictionary and Query Language. The data dictionarydescribes structure of data in the database and also the characteristics of data viz. length, type etc. Thequery language part provides access to data in the database. It also helps generate reports.

The relational structure is the most popular, (particularly with PCs), and is, possibly, the most flexibletype of data organisation for database construction. This type of database is built around the conceptthat data contained in the conventional files can, with some duplication, be transformed into twodimensional tables. Each such table may have any number of rows, but may have only few columns.Data in each column has to be of uniform size, residing a field in a record, for example customers'name.


(a) A Parcel carrying transport company will be having the data on :

(i) Bookings

(ii) Expenses

(iii) Operations

(iv) Accounts

(v) Petty Cash expenses

(vi) Claims

(vii) Complaints

(viii)Delivery

(ix) Maintenance, and so on.

The specimen format is given for booking figures.

Sl. No. Waybill No. Commodity Qty (Unit) Freight Remarks

They will also be collecting the data as on date and cumulative figures for which provisions can bemade. Similarly for each item the format can be suitably devised.

(b) The head office can collect the data branch wise and then collate for all branches and obtain thefigures for the draft and cumulative as required.

Branch Booking for the date Cumulative Booking Cumulative Booking forfor the month the previous month

1

2

3

4

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The software can be suitably devised and the data can be sent by modern/other electronic media dependingupon the available resources. Similarly the other data can be compiled.

(c) If the computers are made available at the branches and the head office the communication can be fastand the uniformity in formats an required data can be standardised.

If the personnel are suitably trained than they will be finding it comfortable to send the requiredinformation on time.

The head office will collate all the branch data quickly and carry out the various analysis and submitto the top management for quick decisions.

Using the software the results can be obtained in Tabular forms and or graphical devices.

When Quantitative Techniques are applied the results are used for quick decisions

Any correction a changes can be introduced and revised results can be obtained.

(d) The following precautionary measures can be taken to prevent possible errors :

(i) The personnel who are involved in collection of data should be sufficiently trained so that thesource data are correct.

(ii) Data entry operators/staff should be trained in entering the data in the computer.

(iii) Source data can be linked to modem/other electronic media to prevent duplication of entry/copy of data.

(iv) The persons similarly who receive the data as the head office should be trained in connectingeach branch and collate the data.

(v) Each computer should be provided with back up in case of failure of electricity.

(vi) Passwords are to be used by these personnel so that there is no problems of whatsoever

(viii)The security for computers, disasters and misuse of computers should be avoided.


(a) Advantages of Centralised Architecture :

(i) Hardware setup for this architecture is less expensive.

(ii) This system provides us all the data related to the entire company. Hence, at any pointof time we can fetch any data from a single point.

(iii) We can get the consolidated MIS report much faster.

(iv) Overall control on operations is much compact

(v) Database administration is centralized, hence tampering is not possible.

Disadvantages of Centralised Architecture :

(i) Network setup is very expensive and its maintenance cost is also very high.

(ii) If the network system breaks down then the entire system will be handicap.

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Advantages of Decentralised Architecture :

(i) It is an efficient and prompt processing of jobs assigned to different processor.

(ii) The system can be easily expanded as per requirement, by addition of more units as required.

(iii) At any particular point we can get, to the point report for a specific data.

(iv) If any network hazard occurs, the system can run smoothly because the operation can be donefrom various points.

(v) Failure of one particular system does not, seriously, affect working of the entire system.

Disadvantages of Decentralised Architecture :

(i) Hardware setup is expensive.

(ii) Consolidate MIS generation is cumbersome

(iii) Database administration is decentralised hence tampering chance is higher.

(b) Some hazards of computerisation are :

(i) Machine breakdown (due to some hardware parts problem)

(ii) Due to virus infection data even can be destroyed

(iii) Due to incompetence of staff the software can be erased from the machine and the data can becorrupted.

(iv) Purposefully any body can damage the system or data can be tampered.

Some of the measures to avoid these hazards are :–

(i) Regular backup of the entire company data

(ii) The computer should be protected from unauthorised user by restrictions on physical entry andit should be password protected, so that, the system can not be started by any unauthorisedperson.

(iii) The second level of password checking should be done by the software itself which is runningfor the computation job.

(iv) Data encryption

(v) Regular virus scanning and virus removal

(vi) Proper method of data recovery and database recreation.

(vii) Contingent planning.

(c) The steps involved in program development are :

(i) Defining objective

(ii) Specifying input and output formats

(iii) Designing program logic

(iv) Selecting programming language

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(v) Coding (writing) program

(vi) Compiling program for detecting syntax errors and logical errors in the source program.

(vii) Testing and debugging of programs

(vii) Documentation.


(a) The techniques for changing over from the old system to new system are :

(i) Direct changeover from the old one to the new one.

(ii) Both systems are running parallely for some time (until and unless the new system isproved as full proof).

(iii) Pilot running.

(iv) Phased Changeover.

(b) Data processing : Data processing is basically concerned with converting raw data into well-orderedinformation as would serve the purpose of the recipient. Data processing activities, in turn, encompassdata input, data manipulation and outputting results of data manipulation.

Components of Computerised Data Processing system :

(i) Hardware: Electronically-operated machines for carrying different operations.

(ii) Software : Step-by-step instructions that tell the computer how to do its work.

(iii) Procedures : These relate to the routines to be followed in such matters as preparation of data,operation of computer, distribution of the processed output and the like.

(iv) Personnel: Computer installations employ the following types of personnel

(a) System Analyst

(b) Programmers

(c) Operators

(v) Data

(c) Controls on computer outputs :

(i) Initial screening of outputs

(ii) Reconciliation of output control totals

(iii) Immediate distribution of outputs

(iv) Systematic manual checks.


(i) Introduction :

• Overview of the company.

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• Operations in various units and offices with future plans

• Current information technology with special emphasis on WWW and internet.

• Financial Accounting and costing system followed.

• Present audit coverage.

New computerised system :

• Hardware, Networking, technologies for data transmission, system software, file structure andprocessing techniques.

• Basic concept of e-mail, Internet, WWW and E-commerce

• Information system designed.

• Codification.

• Input-output documentation

• Data communication system

• Decision Support System

Control and Audit of computer based system :

• Overview.

• Changes in accounting and auditing environment brought about by computers.

• Control system

• Input controls

• Processing controls

• Output controls


• Security controls

Auditing in computer based data processing system :

• Introduction

• Understanding basic features of the computer based system

• Review of application documentation

• Evaluation of controls

• Designing test procedures

• Application of selected audit procedures to critical controls

• Analysis of audit results and reporting

(ii) The guidelines to conduct the audit of computerised accounting system will comprise the followingsteps :

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• Understanding basic features of the computer based system

• Input documentation

• Accounting system

• MIS

• Control system

• Review of application documentation

• Data contents

• Procedures


• System flowchart

• Input formats

• Record layouts

• Coding system

• Exceptions

• Evaluation of control

• Preparation of control checklist

• Appraisal of key activities

• Designing test procedures

• Choosing testing techniques

• Testing processing operations

• Using computers as audit tools

• Test deck

• Generalized audit programme

• Preparation of audit program

• Compliance test

• Evaluation of audit results

• Reporting to management

(iii) Auditing around the computer requires the following steps :

(a) Selection of one or more critical output from the computer system.

(b) Verifications of the results exhibited by the output to ascertain correctness and completeness ofthe transactions processed.

(c) Audit trail to locate the original source of input for verification

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Auditing through the computer implies verification of the computerized system itself and itsefficacy to produce the correct and required output.


End of year Operational Costs Difference Cumulative(in Rs. lakhs) (a–b) difference

Old system New system of operational(a) (b) Cost

1 25 55 –30 –30

2 28 38 –10 –40

3 34 14 –20 –60

4 45 15 30 –30

®X ¬

5 47 16 31 01

Pay-back period lies between 4th and 5th year. Let ‘X’ be the pay-back period, where cumulativeoperational cost is ‘Zero’.

∴ −−

= ++

x 4

5 4

0 30

1 30 [By simple Interpolation Formula]

or, x–4 = 30/31 ; or, x = 4 + 0.97 = 4.97 years = 4 years 11.5 months (approx.)

Hence, Pay back period is 4 years 11.5 months. Hence, the management will recover all of itsinitial cost and start making profit after that period.

(b) Software quality essentially means that it is error-free and satisfies the users needs. Specifically, itmeans Suitability and Maintainability.

Suitability implies reliability, correctness, accuracy, understandability, modifiability, traceability,portability and reusability.

The main tools for Quality Assurance of Software are -

(i) Development, Process itself,

(ii) Reviews,

(iii) Audits,

(iv) Testing and Matrices.

(c) System Changeover:

System Changeover means a process of changing old system to new system. When the programming,creating of database, installation of necessary equipment and user training for the new system iscomplete then, the old system can be replaced by the new system. The method and style of the newsystem will reflect the users response and attitude to the system.

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There are three methods of System Changeover :–

(i) Crash Method : Here the old system will be discontinued and new system will be implemented.

(ii) Parallel Method : Here the old system as well as new system will run in parallel for a perioduntil it works to the satisfaction of all concerned.

(iii) Staged Method : Here the old system is replaced by the new system gradually by stages.


(a) The World Wide Web or www as it is called is a concept based on the internet technology. It is aconcept that provides the technology to navigate the vast resources available on the internet. Theconcepts of hypertext, internet and multimedia are integral to the concept of www. The word ‘web’ inthe www signifies the ability to navigate through the multitude of computers and access texts, graphics,sound files etc. in a web-like fashion.

Internet is the network of hundreds or thousands of computers and computer networks worldwide,which are connected with each other, exchanging information. The network is not controlled by centralauthority or organisation. Instead of data going to a central computer and then to its destination withInternet, the data has many points to go from one computer to another, over a web of computers.

www is a concept while Internet is the physical aspect of it.

(b) Calculation of Record Size :

Field Name Maximum Field Size

Purchase-order-number 6

Vendor code 3

Order-quantity 5

Order-date 6

For record deletion Marker 1

Therefore, Record Size 21

Calculation of File Space Requirement :

Record Size 21

× Maximum expected records 750

= Required Record Space 15750

+ 15% increase in future (15% of 15750) 2363

18113

+ 20% overhead (20% of 18113) 3623

21736

+ 10% contingency (10% of 21736) 2174

Therefore, File Space Required 23910

[For overflow conditions]

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(c) File Retention:

File Retention is the process of permanent storage of data. It can be done into two categories of devices- Random and Sequential. Random devices allow access to information directly and not in the orderin which they are stored. Whereas in sequential devices storage and retrieval has to be done in thesame order.

File Recovery:

It is the process of retrieval of data which has been partially corrupted due to logical or physicalerrors on the storage media. It is done using certain third party utilities.


(a) Data integrity : Concerns about errors, duplicity and inconsistencies of data.

Security : Security refers to the protection of data from unauthorised access and use which coulddamage organisational interest.

Privacy : Privacy refers to unauthorised access and use of data which could damage individualsinterest.

(b) Structured and Unstructured Decisions:

A structured decision is a random one and is made on the basis of certain pre-planned actions. Anunstructured decision is non-routine in nature and is infrequent.

Structured decisions follow certain predetermined logical steps and are programmable. But,unstructured decisions are not amenable to programming.

(c)

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(a)

Type of S/W Advantages Disadvantages

Buying readymade 1. Can be seen and tested 1. May not meet requirements fully2. Will have fewer bugs 2. Enhancements/modifications3. May be less costly may be expensive and time4. Implementation time saving consuming5. Documentation/help facilities 3. Requires development, testing and

implementation time

Built by outside firm 1. Better control over 1. Difficult to negotiate on effort andschedule and cost time required

2. Does not affect day to day 2. Difficult to implement withoutoperations due to development adequate technical knowledgework of the software.

3. Ready skills when new 3. Difficult to maintain afterh/w or s/w platforms implementation

4. Unfamiliarity of the outside partywith business may hamper quality

Built-in-house 1. Familiarity with business 1. Difficult to adhere towill lead to better quality time schedule.

2. Will meet requirements fully 2. Additional manpower cost3. Development of in-house skills 3. Particular skills may not4. May make money by selling be available

the software 4. Improvisation according tosystems followed by others.

(b) Prototyping is the process of quickly building a model of the final application system. It is primarilyused as a means of understanding and communicating the requirements of the users.

Advantages of Prototyping include faster development time, easier end-user learning, betterfunctionality of the software, better communication between users and analysts. The disadvantagesinclude fostering of undue expectations of users, providing users with something that may not getfinally.

The two types of Prototyping approach are (i) Iterative (ii) Throwaway. In iterative type approach,the prototype is developed, demonstrated to users and modified based on users feedback. This processis continued till the prototype evolves into a final system which get implemented.

In throwaway approach, the prototype is only a model of the final system. It is not what is implemented.Once the prototype satisfies the users, a new system is developed based on the same.